> vp; // initialing vactor vp
45 |
46 | for(ll i=1;i<=n;i++)
47 | {
48 | ll k,m=0;
49 | cin>>k;
50 | for(ll j=0;j>x;
53 | m=max(m,x-j+1);
54 | }
55 | vp.push_back({m,k}); // push back pair to vp
56 | }
57 | sort(vp.begin(),vp.end()); // to sort the array
58 | ll p=0,l=0;
59 | for(auto i:vp)
60 | {
61 | p=max(p,i.first-l);
62 | l += i.second;
63 | }
64 | cout<
3 | using namespace std;
4 | int main(){
5 | int n,k;
6 | cin>>n>>k;
7 | //creating the array of size n
8 | int a[n];
9 | for(int i=0;i>a[i];
11 | }
12 | //using the sort function to sorting the array in o(nlogn)
13 | sort(a,a+n);
14 | //checking for k is equal to 0 if k is equal to 0 and first element
15 | //is equal to 1 then print -1 else 1
16 | if(k==0)
17 | cout<<(a[0]==1? -1:1);
18 | else{
19 | //checking for same value of a[k-1] and a[k]
20 | if(a[k-1]==a[k]) cout<<-1;
21 | else cout<
5 | #define N 100005
6 | #define mod 1000000007
7 | #define INF 2147483647
8 | #define ll long long
9 | #define ln "\n"
10 | #define fast() ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);
11 | using namespace std;
12 | int main()
13 | {
14 |
15 | fast();
16 | // d is the number of test cases
17 | ll d;
18 | cin>>d;
19 | while(d--)
20 | {
21 | ll P,a,b,c,x,y;
22 |
23 | // taking the input from the user
24 | cin>>P>>a>>b>>c>>x>>y;
25 |
26 | // total cost of anar will be the product of number of fuljharis and their cost + anar cost
27 | ll Q = x * a + b;
28 |
29 | // total cost of chakri will be the product of number of fuljharis and their cost + chakri cost
30 | ll R = y * a + c;
31 |
32 | ll solution = 0;
33 | if (Q < R) // if chakri is greater then this conditon holds
34 | {
35 | solution = Q;
36 | }
37 | else
38 | {
39 | solution = R; // if annar is greater then this condition holds
40 | }
41 |
42 | cout << P / solution < Fury-MadMax)
48 |
49 |
50 |
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/CodeForces/Nastia and Nearly Good Numbers.cpp:
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1 |
2 | // Problem Link :- https://codeforces.com/problemset/problem/1521/A
3 |
4 | #include
5 | #define N 100005
6 | #define mod 1000000007
7 | #define INF 2147483647
8 | #define ll long long
9 | #define ln "\n"
10 | #define fast() ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);
11 | using namespace std;
12 | int main()
13 | {
14 | fast();
15 |
16 | int c; // c is the number of test cases
17 | cin>>c;
18 | while(c--)
19 | {
20 | // taking the input from the user
21 | ll A,B;
22 | cin>>A>>B;
23 | ll p,q,r;
24 |
25 | // assuming that p is a positive integer that takes a nearly good number.
26 | p=A;
27 |
28 | // for q to be a perfectly good number if it is divisible by the product A*B.
29 | q=A*B;
30 |
31 | // according to question constraint this must hold
32 | r=q+p;
33 |
34 | if(p==q)
35 | {
36 | // if it dont satisfy the conditions print NO
37 | cout<< "NO" < Fury-MadMax)
48 |
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/CodeForces/Permutation_Minimization.cpp:
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1 | //C++ program to find the lexicographically smallest possible sequence of elements in the deque
2 | //after the entire permutation has been processed.
3 | #include
4 | using namespace std;
5 |
6 | int main()
7 | {
8 | int t;
9 | //Taking input for t->number of test cases
10 | cin >> t;
11 | while (t--)
12 | {
13 | int n;
14 | //Taking input for n->permutation size
15 | cin >> n;
16 | vector v(n);
17 |
18 | //Taking inputs for elements of the permutation.
19 | for (int i = 0; i < n; i++)
20 | {
21 | cin >> v[i];
22 | }
23 |
24 | //Creating a deque
25 | deque d;
26 |
27 | //Adding first element in the deque
28 | d.push_back(v[0]);
29 | for(int i=1;iv[i])
32 | {
33 | d.push_front(v[i]); //Adding elements at the beginning if it smaller than the first element
34 | }
35 | else{
36 | d.push_back(v[i]); //Adding elements at the end if it is greater than the first element
37 | }
38 | }
39 |
40 | //Printing the elements of the deque
41 | cout< Master-Helix)
2 |
3 | //question link - https://codeforces.com/contest/1490/problem/C
4 |
5 |
6 | #include
7 | using namespace std;
8 |
9 | #define fast() ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);
10 | #define ll long long
11 | #define l long long
12 |
13 |
14 |
15 | int main()
16 | {
17 | fast();
18 | ll t;
19 | cin>>t;
20 | while(t--) // no. of test cases
21 | {
22 | ll n;
23 | cin>>n; // the number to be expressed as sum of cubes
24 | ll i=1;
25 | int flag=0;
26 | while(i*i*i
4 | using namespace std;
5 | // validating the window for valid numbers of 1, 2 and 3
6 | bool verify( int window[4]){
7 | if(window[0]==0 && window[1]>=1 && window[2]>=1 && window[3]>=1 ) return true;
8 | return false;
9 | }
10 | // function f2 for the slinding window of size mid
11 | bool f2(string s, int mid){
12 | int window[4]={0,0,0,0};
13 | int i=0,p=0;
14 | int n=s.size();
15 | for(;i>t;
46 | while(t--){
47 | string s;
48 | cin>>s;
49 | cout< Master-Helix)
6 |
7 | #include
8 | using namespace std;
9 |
10 | #define fast() ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);
11 | #define ll long long
12 | #define l long long
13 |
14 |
15 |
16 |
17 |
18 | // RAKSHIT
19 | int main()
20 | {
21 |
22 | fast();
23 | ll n;
24 | cin>>n;
25 | vector>v; // store width and height of each rectangle in form of pair inside a vector
26 | for(int i=0;i>w1>>h1;
30 | v.push_back(make_pair(w1,h1));
31 | }
32 |
33 | ll maxi=max(v[0].first,v[0].second); // set maxi as first value of rectangle (maximum of width and height)
34 |
35 | int flag=0;
36 | for(int i=1;imaxi)
41 | {
42 | flag=1;
43 | break;
44 | }
45 | else // if min value is more than previous max , then domino is not formed so break the loop
46 | {
47 | if(val2<=maxi)
48 | {
49 | maxi=val2;
50 | }
51 | else
52 | {
53 | maxi=val;
54 | }
55 | }
56 | }
57 |
58 | if(flag==1) // to print the desired output
59 | {
60 | cout<<"NO";
61 | }
62 | else
63 | {
64 | cout<<"YES";
65 | }
66 | }
67 |
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/CodersArena/Coders Arena.md:
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1 | # Coders Arena
2 |
3 |
4 | 
5 |
6 | # Inspiration behind creating the logo
7 |
8 | I really love creating logos. When I saw the issue regarding *help for logo designing* I knew that I could be of some help to this wonderful community of Sarthak Keshari. So I designed this logo which looks like a cubical arena which symbolizes a place where like minded coders can help each other out. I'm really happy that I get to be a part of this amazing community!
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/CodersArena/coders arena logo.png:
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https://raw.githubusercontent.com/SarthakKeshari/CPP-Questions-and-Solutions/b3097685322caac5124ecfbfd41a221fb6bc8318/CodersArena/coders arena logo.png
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/Coding Competition/Consistency-1.cpp:
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1 | //https://www.facebook.com/codingcompetitions/hacker-cup/2021/qualification-round/problems/A1
2 | #include
3 | using namespace std;
4 | #define ll long long
5 | #define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
6 |
7 | int main(){
8 | fast;
9 | ll t;
10 | cin>>t;
11 | for(int p=1;p<=t;p++){
12 | string s;
13 | cin>>s;
14 | vector v(26);//counter for individual vowel
15 | vector c(26);//counter for individual consonant
16 | ll vc=cc=0;//counter for total vowel and consonant
17 | for(int i=0;i
4 | #include
5 | #include
6 | using namespace std;
7 |
8 | // Function to find the k'th largest element in an array using min-heap
9 | int findKthLargest(vector const &ints, int k)
10 | {
11 | // base case
12 | if (ints.size() < k) {
13 | exit(-1);
14 | }
15 |
16 | // create a min-heap using `std::priority_queue` and insert
17 | // the first `k` array elements into the heap
18 | // `std::greater` is used as the comparison function for min-heap
19 | priority_queue, greater> pq(ints.begin(), ints.begin() + k);
20 |
21 | // do for remaining array elements
22 | for (int i = k; i < ints.size(); i++)
23 | {
24 | // if the current element is more than the root of the heap
25 | if (ints[i] > pq.top())
26 | {
27 | // replace root with the current element
28 | pq.pop();
29 | pq.push(ints[i]);
30 | }
31 | }
32 |
33 | // return the root of min-heap
34 | return pq.top();
35 | }
36 |
37 | int main()
38 | {
39 | vector ints = { 7, 4, 6, 3, 9, 1 };
40 | int k = 2;
41 |
42 | cout << "k'th largest array element is " << findKthLargest(ints, k);
43 |
44 | return 0;
45 | }
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/Data Structure/Linked List/Circular Linked List/Welcome.txt:
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/Data Structure/Linked List/Doubly Linked List/Welcome.txt:
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/Data Structure/Queue/Priority Queue/Welcome.txt:
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/Data Structure/Stack/Next_Greater_Element.cpp:
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1 | // GFG Link to question - https://practice.geeksforgeeks.org/problems/next-larger-element-1587115620/1
2 |
3 | // This Code is Contributed by RAKSHIT PANDEY (Username-----> Master-Helix)
4 |
5 |
6 | // Function to print the Next Greater Element using Stack
7 |
8 | vector nextLargerElement(vector arr, int n){
9 |
10 | vectorv; // to store the output
11 |
12 | v.push_back(-1); // for rightmost element . no right element present
13 | stacks;
14 | s.push(arr[n-1]); //rightmost at top
15 | for(int i=n-2;i>=0;i--) // iterating backwards
16 | {
17 | if(arr[i]
4 | using namespace std;
5 |
6 | struct Node {
7 | int data;
8 | Node* left;
9 | Node* right;
10 |
11 | Node(int d)
12 | {
13 | data = d;
14 | left = NULL;
15 | right = NULL;
16 | }
17 | };
18 |
19 | // Function to calculate height of the tree
20 |
21 | int HeightOfTree(Node* root)
22 | {
23 | if (!root)
24 | return 0;
25 |
26 | int x = HeightOfTree(root->left);
27 | int y = HeightOfTree(root->right);
28 |
29 | return (max(x, y) + 1);
30 | }
31 |
32 | // Function to calculate Diameter of the tree
33 |
34 | int DiameterOfTree(Node* root)
35 | {
36 | if (!root)
37 | return 0;
38 |
39 | int lh = HeightOfTree(root->left); // Height of left subtree
40 | int rh = HeightOfTree(root->right); // Height of right subtree
41 |
42 | int Dia = lh + rh + 1; // Current diameter
43 |
44 | int ld = DiameterOfTree(root->left); // Diameter of left subtree
45 | int rd = DiameterOfTree(root->right); // Diameter of right subtree
46 |
47 | return max(Dia , max(ld, rd));
48 | }
49 |
50 | int main()
51 | {
52 | Node* root = new Node(15);
53 | root->left = new Node(8);
54 | root->right = new Node(3);
55 | root->left->left = new Node(4);
56 | root->left->right = new Node(11);
57 | root->right->left = new Node(10);
58 | root->right->right = new Node(7);
59 | root->left->right->right = new Node(6);
60 | root->right->right->right = new Node(7);
61 | root->left->right->right->left = new Node(14);
62 |
63 | int ans = DiameterOfTree (root);
64 |
65 | cout << ans;
66 |
67 | return 0;
68 | }
69 |
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/Data Structure/Trees/Binary Trees/Height_of_Tree.cpp:
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1 | // 3. Calculate Height of a Binary tree.
2 |
3 | #include
4 | using namespace std;
5 |
6 | struct Node {
7 | int data;
8 | Node* left;
9 | Node* right;
10 |
11 | Node(int d)
12 | {
13 | data = d;
14 | left = NULL;
15 | right = NULL;
16 | }
17 | };
18 |
19 | // Function to calculate height of the tree
20 |
21 | int HeightOfTree(Node* root)
22 | {
23 | if (!root)
24 | return 0;
25 |
26 | int x = HeightOfTree(root->left); // Height of left subtree
27 | int y = HeightOfTree(root->right); // Height of right subtree
28 |
29 | return (max(x, y) + 1);
30 | }
31 |
32 | int main()
33 | {
34 | Node* root = new Node(15);
35 | root->left = new Node(8);
36 | root->right = new Node(3);
37 | root->left->left = new Node(4);
38 | root->left->right = new Node(11);
39 | root->right->left = new Node(10);
40 | root->right->right = new Node(7);
41 | root->left->right->right = new Node(6);
42 | root->right->right->right = new Node(7);
43 |
44 | int ans = HeightOfTree(root);
45 |
46 | cout << ans;
47 |
48 | return 0;
49 | }
50 |
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/Data Structure/Trees/Binary Trees/No_of_Nodes.cpp:
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1 | // 1. Calculate number of nodes in a Binary Tree
2 |
3 | #include
4 | using namespace std;
5 |
6 | struct Node {
7 | int data;
8 | Node* left;
9 | Node* right;
10 |
11 | Node(int d)
12 | {
13 | data = d;
14 | left = NULL;
15 | right = NULL;
16 | }
17 | };
18 |
19 | // Recursive function to count the nodes
20 |
21 | int CountNodes(Node* root)
22 | {
23 | if (!root)
24 | return 0;
25 |
26 | int a = CountNodes(root->left); // Nodes of left subtree
27 | int b = CountNodes(root->right); // Nodes of right subtree
28 |
29 | return a + b + 1;
30 | }
31 |
32 | int main()
33 | {
34 | Node* root = new Node(15);
35 | root->left = new Node(8);
36 | root->right = new Node(3);
37 | root->left->left = new Node(4);
38 | root->left->right = new Node(11);
39 | root->right->left = new Node(10);
40 | root->right->right = new Node(7);
41 | root->left->right->right = new Node(6);
42 | root->right->right->right = new Node(7);
43 |
44 | int ans = CountNodes(root);
45 |
46 | cout << ans;
47 |
48 | return 0;
49 | }
50 |
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/Data Structure/Trees/Binary Trees/Sum_of_Nodes.cpp:
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1 | // 2. Calculate the sum of nodes of a binary tree
2 |
3 | #include
4 | using namespace std;
5 |
6 | struct Node {
7 | int data;
8 | Node* left;
9 | Node* right;
10 |
11 | Node(int d)
12 | {
13 | data = d;
14 | left = NULL;
15 | right = NULL;
16 | }
17 | };
18 |
19 | int SumOfNodes(Node* root)
20 | {
21 | if (!root)
22 | return 0;
23 |
24 | int a = SumOfNodes(root->left); // Sum of left subtree
25 | int b = SumOfNodes(root->right); // Sum of right subtree
26 |
27 | return a + b + root->data;
28 | }
29 |
30 | int main()
31 | {
32 | Node* root = new Node(15);
33 | root->left = new Node(8);
34 | root->right = new Node(3);
35 | root->left->left = new Node(4);
36 | root->left->right = new Node(11);
37 | root->right->left = new Node(10);
38 | root->right->right = new Node(7);
39 | root->left->right->right = new Node(6);
40 | root->right->right->right = new Node(7);
41 |
42 | int ans = SumOfNodes(root);
43 |
44 | cout << ans;
45 |
46 | return 0;
47 | }
48 |
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/Data Structure/Trees/Red Black Trees/Welcome.txt:
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/Datatype-String/string.txt:
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1 | Main Topic – Data Types
2 | Sub-topic – string
3 |
4 | Formal Definition
5 | • It is a non-primitive datatype.
6 | • It is used for storing sequence of characters enclosed within double quotes.
7 | • One must include library, header file in the source code to use strings.
8 | Easy and Clear Explanation (along with examples)
9 | string s = “Happy Birthday”
10 | Here we have taken s as a string datatype which stores Happy Birthday as sequence of characters within double quotes.
11 | Solved Questions
12 | Write a program to print a string.
13 |
14 | #include
15 | #include
16 | using namespace std;
17 | int main()
18 | {
19 | string s = “Feeling Cool”;
20 | cout << s;
21 | return 0;
22 | }
23 |
24 | Output
25 |
26 | Feeling Cool
27 | Practice Questions
28 | Write a program to check whether it is a palindrome string or not.
29 | Write a program to multiply 2 strings.
30 |
31 | -Anushka Srivastava
32 |
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/Datatype-double/double.txt:
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1 |
2 | Main Topic – Data Types
3 | Sub-topic - double
4 |
5 | Formal Definition
6 | • It is a primitive datatype.
7 | • It is used to store decimal values.
8 | • The size of double is 8 bytes i.e. 64 bits.
9 | • The range of double is 1.7e-308 to 1.7e+308
10 |
11 |
12 | Easy and Clear Explanation (along with examples)
13 | Example
14 |
15 | double n = 22.3
16 |
17 | Here we have taken n as a variable of double datatype which store decimal values like 22.3 .
18 |
19 |
20 | Solved Questions(Maximum 2)
21 |
22 | Write a program to subtract to decimal numbers.
23 | #include
24 | using namespace std;
25 | int main()
26 | {
27 | double n1, n2, sub;
28 | n1 = 22.4;
29 | n2 = 12.2;
30 | sub = n1 – n2;
31 |
32 | cout<< “Subtraction of two decimal numbers : “ << sub << endl;
33 | return 0;
34 | }
35 |
36 | Output
37 | Subtraction of two decimal numbers : 10.2
38 |
39 | Write a program to add two decimal numbers.
40 | #include
41 | using namespace std;
42 | int main()
43 | {
44 | double n1,n2, sum;
45 | n1 = 3.6;
46 | n2 = 2.2
47 | sum = n1 + n2;
48 |
49 | cout<< “Sum of two decimal numbers : “ << sum << endl;
50 | return 0;
51 | }
52 | Output
53 | Sum of two decimal numbers : 5.8
54 |
55 | Practice Questions(Minimum 2)
56 | *Medium to Hard level - This line is for developers info. Kindly remove it before creating a PR
57 | Write a program to take base and height as inputs from the user and find the area of triangle.
58 | Formula – 1/2 * base * height
59 | Write a program to find the sum of the n terms of a series.
60 | 1/2 + 3/4 + 5/6 + 7/8 + 9/10 --------------- n terms
61 |
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/Datatype-int/int.txt:
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1 | Main Topic – Data Types
2 | Sub-topic - int
3 |
4 |
5 | Formal Definition
6 | • It is a primitive datatype.
7 | • It stores integer values.
8 | • The size of int is 4 bytes i.e. 32 bits.
9 | • The range is -2147483648 (-2^31) to 2147483647(2^31-1).
10 |
11 |
12 | Easy and Clear Explanation (along with examples)
13 | Example
14 | Declaring a value
15 | int a = 2230
16 | Here we have taken a is a variable of int type which store integer values like 2230.
17 |
18 | Solved Questions(Maxiumum 2)
19 |
20 | Write a program to add to integer numbers.
21 | public class Solution
22 | {
23 | public static void main(String[] args) {
24 | int n1 = 10;
25 | int n2 = 20;
26 | int sum = n1+n2;
27 | System.out.println("Sum of two integer number : "+sum);
28 | }
29 | }
30 |
31 | Output
32 | Sum of two integer number : 30
33 |
34 | Write a program to multiply two integer numbers.
35 | public class Solution
36 | {
37 | public static void main(String[] args) {
38 | int n1 = 2;
39 | int n2 = 3;
40 | int mul = n1*n2;
41 | System.out.println("Multiplication of two integer numbers : "+mul);
42 | }
43 | }
44 |
45 | Output
46 | Multiplication of two integer numbers : 6
47 |
48 | Practice Questions(Minimum 2)
49 | *Medium to Hard level - This line is for developers info. Kindly remove it before creating a PR
50 | • Write a program to check whether a number is a Palindrome number or not?
51 | A palindromic number is a number that remains the same when its digits are reversed.
52 | E.g. 121
53 | • Write a program to check whether a number is a Armstrong number or not?
54 | Armstrong number is a number that is equal to the sum of cubes of its digits.
55 | E.g. 153
56 |
57 | -Anushka Srivastava
58 |
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/Documentation/Installing IDE- Eclipse for C++/Steps to install Eclipse for C++.txt:
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1 | Steps to install Eclipse for C++
2 | ----------------------------------------------------------------------------------------------------
3 |
4 |
5 | Pre-requisites:
6 | -system with processor intel i3 or ryzen 3 and above
7 | -ram: 4gb and above
8 | -internet connectivity to download eclipse and some packages
9 | ----------------------------------------------------------------------------------------------------------------------
10 | step 1: Download ans Install Mingw (https://sourceforge.net/projects/mingw/)
11 | Eclipse CDT uses C/C++ Compiler. Hence before we can start using Eclipse CDT for C/C++ development,
12 | we need to have a proper GCC compiler on our system. We can have ‘MinGW’ compiler on our machine that will be used by eclipse.
13 | img4
14 | step 2: add it to environment path variable
15 | img1
16 | Step 3: download eclipse (https://www.eclipse.org/downloads/)
17 | img2
18 | Step 4: install eclipse
19 | img3
20 | Step 5: open eclipse
21 | img5
22 | Step 6:click on files then click on c++ project
23 | img8
24 | Step 7: give the name to project(cpp program)
25 | -Here you can specify the name of the project. You can select an Empty project or a sample “Hello World” application project.
26 | img9
27 | Step 8: write a program
28 | img10
29 | Step 9: click on green play button to execute the code (img11)
30 | In this tutorial, we have seen the features, installation, configuration, and development using Eclipse CDT IDE.
31 | Although Eclipse IDE is primarily used for Java development, we can also use it for development using other programming languages like C/C++, PHP, Perl, Python to name a few.
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1 | Steps to install Eclipse for C++
2 | ----------------------------------------------------------------------------------------------------
3 |
4 |
5 | Pre-requisites:
6 | -system with processor intel i3 or ryzen 3 and above
7 | -ram: 4gb and above
8 | -internet connectivity to download eclipse and some packages
9 | ----------------------------------------------------------------------------------------------------------------------
10 | step 1: Download ans Install Mingw (https://sourceforge.net/projects/mingw/)
11 | Eclipse CDT uses C/C++ Compiler. Hence before we can start using Eclipse CDT for C/C++ development,
12 | we need to have a proper GCC compiler on our system. We can have ‘MinGW’ compiler on our machine that will be used by eclipse.
13 |
14 | step 2: add it to environment path variable
15 |
16 | Step 3: download codeblocks
17 | img(1),img(2)
18 | Step 4: install codeblocks
19 | img(3),img(4),img(5)
20 | Step 5: open codeblocks
21 | img(7)
22 | Step 6:click on files then click on c++ project
23 | img(9)
24 | Step 7: give the name to project(cpp program)
25 | -Here you can specify the name of the project. You can select an Empty project or a sample “Hello World” application project.
26 | img(14)
27 | Step 8: write a program
28 | img(14)
29 | Step 9: click on yes button to execute the code (img(15))
30 | In this tutorial, we have seen the features, installation, configuration, and development using CodeBlocks IDE.
31 | we can also use it for development using other programming languages like C/C++, PHP, Perl, Python to name a few.
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26 |
27 |
28 | ### Practice Questions(Minimum 2)
29 | #### *Medium to Hard level - This line is for developers info. Kindly remove it before creating a PR
30 | - Question 1
31 | - Question 2
32 |
33 |
34 |
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/Dynamic Programming/Editdistance.cpp:
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1 | //Given two strings str1 and str2 and perform add, remove, replace operations that can performed on str1. Find minimum number of operations required to convert ‘str1’ into ‘str2’
2 | #include
3 | #include
4 |
5 | using namespace std;
6 |
7 | int min(int x,int y,int z) //Function to Calculate the minimum of three number
8 | {
9 | return min(min(x,y),z); //Return minimum of three number
10 | }
11 |
12 | int editdistance(string s1,string s2,int n,int m)
13 | {
14 | if(n==0) //If length of string1 is zero then return lenght of string2
15 | {
16 | return m;
17 | }
18 | if(m==0)
19 | {
20 | return n; //If length of string2 is zero then return lenght of string1
21 | }
22 | if(s1[n-1]==s2[m-1])
23 | {
24 | return editdistance(s1,s2,n-1,m-1); //If last char of both string is same the return digonal element
25 | }
26 | else{
27 | return 1+min(editdistance(s1,s2,n-1,m),editdistance(s1,s2,n,m-1),editdistance(s1,s2,n-1,m-1)); //Else return min of ((x-1,y-1),(x,y-1),(x-1,y)) +1
28 | }
29 |
30 | }
31 | //Driver code
32 | int main()
33 | {
34 | string s1;
35 | cout<<"Enter the string 1"<>s1;
37 |
38 | string s2;
39 | cout<<"Enter the string 2"<>s2;
41 | int n=s1.size();
42 | int m=s2.size();
43 |
44 | cout<<"Total operations to convert string 1 to string 2 IS "<
3 | using namespace std;
4 |
5 |
6 | int max(int a, int b) //Function to calculate the maximum
7 | {
8 | if(a>b)
9 | {
10 | return a;
11 | }
12 | else
13 | {
14 | return b;
15 | }
16 | }
17 |
18 | int eggdrop(int n, int k)
19 | {
20 | if (k == 1 || k == 0) //If Number of floor is 1 or 0 then return number of florr
21 | return k;
22 |
23 | if (n == 1) //If the number of egg is 1 then return number of floors
24 | return k;
25 |
26 | int min = INT_MAX, res;
27 |
28 | for (int i = 1; i <= k; i++) {
29 | res = max(eggdrop(n - 1, i - 1),eggdrop(n, k - i)); //If egg breaks then floor will k-1 and egg will be n-1
30 | if (res < min) //Else floor will be k-i and number of eggs will be same
31 | min = res;
32 | }
33 |
34 | return min + 1;
35 | }
36 |
37 | int main()
38 | {
39 | int n;
40 | cout<<"Enter the number of Eggs "<>n;
42 | int k;
43 | cout<<"Enter the number of Floors "<>k;
45 |
46 | cout<
3 | #include
4 | using namespace std;
5 |
6 | int lps(char*seq,int i,int j)
7 | {
8 | if(i==j) //If string has only one value then return 1;
9 | {
10 | return 1;
11 | }
12 |
13 | if(seq[i]==seq[j] && i+1==j) //If string has only one value then return 2;
14 | {
15 | return 2;
16 | }
17 |
18 | if(seq[i]==seq[j]) //If string has only first and last character are same;
19 | {
20 | return lps(seq,i+1,j-1)+2;
21 | }
22 | else{
23 | return max(lps(seq,i+1,j),lps(seq,i,j-1)); //If string has only first and last character are not same then call the recursive function;
24 | }
25 |
26 |
27 | }
28 |
29 | //Driver class
30 | int main()
31 | {
32 | int n;
33 | cout<<"Enter the number of characters"<>n;
35 | char seq[n];
36 | cout<<"Enter the String"<>seq;
38 |
39 | cout<<"longest palindromic subsequence is "<
9 | using namespace std;
10 |
11 | // Preprocessing helps the code run faster
12 | #define fl(i, a, b) for (int i = a; i < b; i++)
13 |
14 | // Function that return
15 | int maxProfit(int* prices, int size)
16 | {
17 | // maxProfit adds up the difference between
18 | // adjacent elements if they are in increaisng order
19 | int maxProfit = 0;
20 | // The loop starts from 1
21 | // as its comparing with the previous
22 | fl(i, 1, size) if (prices[i] > prices[i - 1]) maxProfit
23 | += prices[i] - prices[i - 1];
24 | return maxProfit;
25 | }
26 |
27 | // Driver Function
28 | int main()
29 | {
30 | int prices[] = { 100, 180, 260, 310, 40, 535, 695 };
31 | int N = sizeof(prices) / sizeof(prices[0]);
32 | cout << maxProfit(prices, N) << endl;
33 | return 0;
34 | }
35 | // This code is contributed by Rajat Jain
36 |
37 |
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/GeeksForGeeks/Cuttingrod.cpp:
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1 | //Given a rod of length n inches and an array of prices that includes prices of all pieces of size smaller than n. Determine the maximum value obtainable by cutting up the rod and selling the pieces.
2 | #include
3 | #include
4 |
5 | using namespace std;
6 |
7 | //Function to find the maximum of two integers
8 | int max(int a, int b){
9 | if(a>b)
10 | {
11 | return a;
12 | }
13 | else{
14 | return b;
15 | }
16 | }
17 |
18 | //Function to find the maximum profit
19 | int cutRod(int price[], int n)
20 | {
21 | int maxprofit;
22 | int arr[n+1]; //Let an array of size n+1 to store different profit possible.
23 | arr[0] = 0; //Intilaise the first element of let array be 0
24 |
25 |
26 | for (int i = 1; i<=n; i++)
27 | {
28 | maxprofit = INT_MIN;
29 | for (int j = 0; j < i; j++)
30 | {
31 | maxprofit = max(maxprofit, price[j] + arr[i-j-1]); //Find the max profit possible by checking all the possible cases
32 |
33 | }
34 |
35 | arr[i] = maxprofit; //Store all the profit in acsending order in array
36 | }
37 |
38 | return arr[n]; //return the last element of array
39 | }
40 |
41 | //Driver code
42 | int main()
43 | {
44 | int n;
45 | cout<<"Enter the length of Rod "<>n;
47 | int profit[n];
48 | cout<<"Enter the profit of different pieces of Rod "<>profit[i]; //Input the profit of different pieces of Rod
52 | }
53 | cout <<"Maximum profit "<
3 | using namespace std;
4 | // } Driver Code Ends
5 | class Solution
6 | {
7 | public:
8 | // arr: input array
9 | // n: size of array
10 | //Function to find the sum of contiguous subarray with maximum sum.
11 | int maxSubarraySum(int arr[], int n)
12 | {
13 | int curr_sum = arr[0];
14 | int max_sum = arr[0];
15 | for (int i = 1; i < n; i++)
16 | {
17 | curr_sum = max(curr_sum + arr[i], arr[i]);
18 | max_sum = max(max_sum, curr_sum);
19 | }
20 | // Your code here
21 | return max_sum;
22 | }
23 | };
24 | // { Driver Code Starts.
25 | int main()
26 | {
27 | int t, n;
28 | cin >> t; //input testcases
29 | while (t--) //while testcases exist
30 | {
31 | cin >> n; //input size of array
32 | int a[n];
33 | for (int i = 0; i < n; i++)
34 | cin >> a[i]; //inputting elements of array
35 | Solution ob;
36 | cout << ob.maxSubarraySum(a, n) << endl;
37 | }
38 | }
39 | // } Driver Code Ends
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/GeeksForGeeks/Median_in_sorted_rowwise_matrix/Question.md:
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1 | ## Question Statement:
2 | Given a row wise sorted matrix of size RxC where R and C are always odd, find the median of the matrix.
3 |
4 | ## Exmaples:
5 | Example 1:
6 |
7 | Input:
8 | R = 3, C = 3
9 | M = [[1, 3, 5],
10 | [2, 6, 9],
11 | [3, 6, 9]]
12 |
13 | Output: 5
14 |
15 | Explanation:
16 | Sorting matrix elements gives us
17 | {1,2,3,3,5,6,6,9,9}. Hence, 5 is median.
18 |
19 |
20 | Example 2:
21 |
22 | Input:
23 | R = 3, C = 1
24 | M = [[1], [2], [3]]
25 | Output: 2
26 |
27 |
28 | ### Expected Time Complexity: O(32 * R * log(C))
29 | ### Expected Auxiliary Space: O(1)
30 |
31 |
32 | ## Constraints:
33 | - 1<= R,C <=150
34 | - 1<= matrix[i][j] <=2000
35 |
36 | ## Probelm link:
37 | https://practice.geeksforgeeks.org/problems/median-in-a-row-wise-sorted-matrix1527/1
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/GeeksForGeeks/Median_sorted_array_diff_size/Question.md:
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1 | https://practice.geeksforgeeks.org/problems/median-of-2-sorted-arrays-of-different-sizes/1#
2 |
Given two sorted arrays array1 and array2 of size m and n respectively. Find the median of the two sorted arrays.
3 |
4 |
Example 1:
5 |
6 |
Input:
7 |
m = 3, n = 4
8 |
array1[] = {1,5,9}
9 |
array2[] = {2,3,6,7}
10 |
Output: 5
11 |
Explanation: The middle element for
12 |
{1,2,3,5,6,7,9} is 5
13 |
Example 2:
14 |
15 |
Input:
16 |
m = 2, n = 4
17 |
array1[] = {4,6}
18 |
array2[] = {1,2,3,5}
19 |
Output: 3.5
20 |
Your Task:
21 |
The task is to complete the function MedianOfArrays() that takes array1 and array2 as input and returns their median.
22 |
23 |
Can you solve the problem in expected time complexity?
24 |
25 |
Expected Time Complexity: O(min(log n, log m)).
26 |
Expected Auxiliary Space: O((n+m)/2).
27 |
28 |
Constraints:
29 |
0 ≤ m,n ≤ 10^4
30 |
1 ≤ array1[i], array2[i] ≤ 10^5
31 |
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/GeeksForGeeks/Minimum-Platform.cpp:
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1 | // Program to find minimum number of platforms
2 | // required on a railway station
3 | #include
4 | using namespace std;
5 |
6 |
7 | class Solution{
8 | public:
9 | //Function to find the minimum number of platforms required at the
10 | //railway station such that no train waits.
11 | int findPlatform(int arr[], int dep[], int n)
12 | {
13 | // Your code here
14 | sort(arr,arr+n); //Sort the Array of Arrival Time
15 | sort(dep,dep+n); //Sort the Array of Departure Time
16 | int i=1,j=0; //Two pointer one for arr and other for dep array
17 | int plat=1,result=1; //plat contains platform needed and result will contian the final answer to the problem
18 | while(i>t;
42 | while(t--)
43 | {
44 | int n;
45 | cin>>n;
46 | int arr[n];
47 | int dep[n];
48 | for(int i=0;i>arr[i];
50 | for(int j=0;j>dep[j];
52 | }
53 | Solution ob;
54 | cout <
4 | using namespace std;
5 |
6 |
7 | // Function to find the maximum increasing sum
8 | int maxincsum(int arr[], int n)
9 | {
10 | int max = 0;
11 | int mis[n]; //let an array of size n
12 |
13 |
14 | for (int i = 0; i < n; i++ )
15 | {
16 | mis[i] = arr[i]; //Store given array in mis array
17 | }
18 |
19 |
20 |
21 | for (int i = 1; i < n; i++ )
22 | for (int j = 0; j < i; j++ )
23 | if (arr[i] > arr[j] && mis[i] < mis[j] + arr[i]) //If array is increasing and sum is increasing
24 | {
25 | mis[i] = mis[j] + arr[i]; //Add the value in mis array
26 | }
27 |
28 |
29 | for (int i = 0; i < n; i++ ) //find the max sum in array
30 | {
31 | if ( max < mis[i] ){
32 | max = mis[i];
33 |
34 | }
35 |
36 | }
37 |
38 | return max; //Return the maximum sum
39 | }
40 |
41 | // Driver Code
42 | int main()
43 | {
44 | int n;
45 | cout<<"Enter the size of array "<>n; //Input the size of array;
47 | int arr[n];
48 | cout<<"Enter the elements of array "<>arr[i]; //Input the elements of array;
52 | }
53 |
54 | cout << "Sum of maximum sum increasing " <
5 | using namespace std;
6 | int main()
7 | {
8 | int a[50],b[50],i,j,N;
9 | cout<<"How many elements you want to enter:-";
10 | cin>>N; // inputting number of elements
11 | cout<<"Enter any "<>a[i]; //inputting the elements that we want to swap
15 | }
16 | cout<<"Reverse of Array:- ";
17 |
18 | for(i=N-1,j=0; i>=0;i--,j++)
19 | {
20 | b[i]=a[j]; //swapping the elements
21 | }
22 | for(i=0; i
11 | #include
12 | using namespace std;
13 |
14 | int main() {
15 | int i;
16 | long l;
17 | char c;
18 | float f;
19 | double d;
20 | scanf("%d %ld %c %f %lf", &i, &l, &c, &f, &d);
21 |
22 | printf("%d\n%ld\n%c\n%f\n%lf", i, l, c, f, d);
23 | return 0;
24 | }
25 |
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/Hackerrank/Birthday_Cake_Candles.cpp:
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1 | /*
2 |
3 | You are in charge of the cake for a child's birthday. You have decided the cake will have one candle for each year of their total age.
4 | They will only be able to blow out the tallest of the candles. Count how many candles are tallest.
5 |
6 | https://www.hackerrank.com/challenges/birthday-cake-candles/problem
7 |
8 | */
9 |
10 | #include
11 |
12 | using namespace std;
13 |
14 | int main()
15 | {
16 | int n,temp=0,count=0,i;
17 | cin>>n;
18 | vector array(n);
19 | for(i=0;i>array[i];
22 | }
23 |
24 | sort(array.begin(),array.end()); //rearrange the elements in ascending order
25 | temp=array[n-1];
26 |
27 | for(i=0;i
2 | #include
3 |
4 | using namespace std;
5 |
6 | int main(){
7 | int y;
8 | cin >> y;
9 | string s = "12.09."; //as Programmer's day will always be in the month of september, we create a string already containing date 12 and month September
10 |
11 | if (y < 1918){
12 | if (y%4 == 0){ //before 1918, leap year was considered only when the year was divisible by 4
13 | s[0] = '1'; //so the 256th day on leap year will be 12.09.yyyy, hence keeping the string as it is
14 | s[1] = '2';
15 | }
16 | else {
17 | s[0] = '1'; //the 256th day on a non-leap year will fall on the 13.09.yyyy
18 | s[1] = '3';
19 | }
20 | }
21 | else if (y == 1918){
22 | s[0] = '2';
23 | s[1] = '6';
24 | //in year 1918, the days 1-13th Feb are not counted, so we'll have to add 13 to 13th Sep, which is 26.09.1918
25 | }
26 | else {
27 | if (y%400 == 0 || (y%4 == 0) && (y%100 != 0)){ //after 1918, the condition for leap year becomes that it should be divisible by 400 or should be divisible by 4 and not by 100
28 | s[0] = '1'; //so the 256th day on leap year will be 12.09.yyyy, hence keeping the string as it is
29 | s[1] = '2';
30 | }
31 | else{
32 | s[0] = '1'; //the 256th day on a non-leap year will fall on the 13.09.yyyy
33 | s[1] = '3';
34 | }
35 | }
36 | cout << s << y << endl; //appending the year to the string
37 |
38 | return 0;
39 | }
40 |
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/Hackerrank/For Loop/For Loop.cpp:
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1 | // QUESTION:-
2 |
3 | // A for loop is a programming language statement which allows code to be repeatedly executed.
4 |
5 | // The syntax is
6 |
7 | // for ( ; ; )
8 | //
9 | // expression_1 is used for intializing variables which are generally used for controlling the terminating flag for the loop.
10 | // expression_2 is used to check for the terminating condition. If this evaluates to false, then the loop is terminated.
11 | // expression_3 is generally used to update the flags/variables.
12 | // A sample loop is
13 |
14 | // for(int i = 0; i < 10; i++) {
15 | // ...
16 | // }
17 | // In this challenge, you will use a for loop to increment a variable through a range.
18 |
19 |
20 |
21 | // Logic:- You will be given two positive integers, a and b (a<=b), separated by a newline.
22 |
23 | // For each integer n, in the interval [a,b]:
24 | // If 1<=n<=9 then print the English representation of it in lowercase. That is "one" for 1, "two" for 2, and so on.
25 | // Else if n>9 and it is an even number, then print "even".
26 | // Else if n>9 and it is an odd number, then print "odd".
27 |
28 |
29 |
30 |
31 | // SOLUTION:- For loop based.
32 |
33 |
34 | #include
35 | using namespace std;
36 | int main()
37 | { // Variable a,b
38 | int a,b;
39 | string c[]={"","one","two","three","four","five","six","seven","eight","nine"};
40 | cin>>a>>b; //Here cpp takes users value and run it.
41 | for(int i=a;i<=b;i++) // Here for loop starts.
42 | cout<<((i<=9)?c[i]:((i%2==0)?"even":"odd"))<
5 | #include
6 | #include
7 | #include
8 | #include
9 | using namespace std;
10 |
11 | int arr[7][7];
12 | int sum(int stx , int sty) //add up each element from subarrays for a square of 3 * 3.
13 | {
14 | return arr[stx][sty] + arr[stx][sty+1] + arr[stx][sty+2] + arr[stx+1][sty+1] + arr[stx+2][sty] + arr[stx+2][sty+1] + arr[stx+2][sty+2];
15 | }
16 | int main() {
17 | int ans = -100;
18 | for(int i = 0 ; i < 6 ; i++){ // Taking the input for the matrix
19 | for(int j = 0 ; j < 6 ; j++){
20 | cin >> arr[i][j];
21 | }
22 | }
23 |
24 | for(int i = 0 ; i < 4 ; i++){ // first for loop to iterate over the array.
25 | for(int j = 0 ; j < 4 ; j++){ // second for loop to iterate over subarray.
26 | ans = max(ans , sum(i,j)); // ans variable that will hold highest sum of the hourglass.
27 | }
28 | }
29 |
30 | cout << ans << endl;
31 | return 0;
32 | }
33 |
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/Hackerrank/Variable-sized-arrays/question.md:
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1 | ## Variable Sized Arrays
2 |
3 |
4 | - link --
5 | [Variable sized arrays](https://www.hackerrank.com/challenges/variable-sized-arrays/problem)
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/Hackerrank/Variable-sized-arrays/solution.cpp:
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1 | //solution
2 |
3 | #include
4 | #include
5 | #include
6 | #include
7 | #include
8 | using namespace std;
9 |
10 |
11 | int main() {
12 |
13 | int noofarr,noofqueries; //initializing no:of arr variable and no:of queries
14 | cin>>noofarr>>noofqueries; //getting input
15 | vector>bigarr; //declaring an vector array which contains vector arrays as elements
16 | for(int i=0;i>arrsize;
19 | vectorsmallarr;
20 | for(int j=0;j>ele;
22 | smallarr.push_back(ele);
23 | }
24 | bigarr.push_back(smallarr);
25 |
26 | }
27 | for(int k=0;k>arrindex>>eleindex;
30 | cout< Solution::findSubstring(string A, const vector &B) {
11 | unordered_map map;
12 | unordered_map tempMap;
13 | int n = B.size();
14 | vector result;
15 | if(n == 0) return result;
16 |
17 | int m = B[0].length();
18 | if(A.length() < m*n) return result;
19 |
20 | for(int i=0;i>& dungeon) {
4 | vector> dp(dungeon.size(), vector(dungeon[0].size(),-1)); //dp value is minimum initial health to get to the end from i, j
5 | return DFS(dungeon,0,0,dp);
6 | }
7 |
8 | int DFS(vector>& dungeon, int i, int j, vector>& dp){
9 | if (i == dungeon.size()-1 && j == dungeon[0].size()-1) //If we are on the last cell return the health needed to pass the cost if negative, otherwise return 1 if positive
10 | return dungeon[i][j] > 0? 1 : 1 - dungeon[i][j];
11 | if (dp[i][j] != -1) //If we already calculated the cell previously, use the value
12 | return dp[i][j];
13 |
14 | if (i == dungeon.size()-1) //If in the last row, only DFS with j increments
15 | return dp[i][j] = max(1, DFS(dungeon,i,j+1,dp)-dungeon[i][j]);
16 | if (j == dungeon[0].size()-1) //If in the last column, onlly DFS with i increments
17 | return dp[i][j] = max(1, DFS(dungeon,i+1,j,dp)-dungeon[i][j]);
18 |
19 | //If not in the last row or column, return minimum of lateral or vertical step-DFS
20 | return dp[i][j] = max(1, min(DFS(dungeon,i+1,j,dp)-dungeon[i][j],DFS(dungeon,i,j+1,dp)-dungeon[i][j]));
21 | }
22 | };
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/LeetCode/ Dungeon Game/readme.md:
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1 | Question Link - https://leetcode.com/problems/dungeon-game/
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/LeetCode/3Sum/question.txt:
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1 | Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
2 |
3 | Notice that the solution set must not contain duplicate triplets.
4 |
5 |
6 |
7 | Example 1:
8 | Input: nums = [-1,0,1,2,-1,-4]
9 | Output: [[-1,-1,2],[-1,0,1]]
10 |
11 | Example 2:
12 | Input: nums = []
13 | Output: []
14 |
15 | Example 3:
16 | Input: nums = [0]
17 | Output: []
18 |
19 |
20 | Constraints:
21 | 0 <= nums.length <= 3000
22 | -10^5 <= nums[i] <= 10^5
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/LeetCode/3Sum/readme.md:
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1 | Question
2 |
3 | Link - [3Sum](https://leetcode.com/problems/3sum)
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/LeetCode/Check If It Is a Straight Line.cpp:
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1 | //1232. Check If It Is a Straight Line
2 | //https://leetcode.com/problems/check-if-it-is-a-straight-line/
3 |
4 | class Solution {
5 | public:
6 | bool checkStraightLine(vector>& coordinates) {
7 | //if points are two then return true
8 | if(coordinates.size()==2)
9 | return true;
10 | else{
11 | int x1=coordinates[0][0],x2=coordinates[1][0],y1=coordinates[0] [1],y2=coordinates[1][1];
12 | for(int i=2;i
8 | using namespace std;
9 | int main(){
10 |
11 | // n = no. of stairs
12 | int n;
13 | cin>>n;
14 |
15 | int prev = 0;
16 | int cur = 1;
17 | int temp;
18 |
19 | for (int i = 0; i < n; i++) {
20 | temp = prev;
21 | prev = cur;
22 | cur += temp;
23 | }
24 | cout< findDuplicates(vector& nums) {
4 | vector v;
5 | int n=nums.size(),k;
6 | for(int i=0;i& gain) {
6 | //maximum altitude
7 | int ans = 0;
8 | //current altitude
9 | int altitude = 0;
10 |
11 | for(int i: gain){
12 | altitude += i;
13 | //it checks if altitude is greater than max altitude(i.e., ans) if so, then variable ans is updated.
14 | ans = max(ans, altitude);
15 | }
16 |
17 | return ans;
18 | }
19 | };
20 |
--------------------------------------------------------------------------------
/LeetCode/Fizz Buzz.cpp:
--------------------------------------------------------------------------------
1 | class Solution {
2 | public:
3 | vector fizzBuzz(int n) {
4 | vector str;
5 | //first check for divisiblity by 3 and 5 both then 3 and 5
6 | for(int i=1;i<=n;i++){
7 | if(i%3==0 and i%5==0)
8 | str.push_back("FizzBuzz");
9 | else if(i%3==0)
10 | str.push_back("Fizz");
11 | else if(i%5==0)
12 | str.push_back("Buzz");
13 | else
14 | str.push_back(to_string(i));
15 | }
16 | return str;
17 | }
18 | };
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/LeetCode/GenerateParentheses/generateparentheses.cpp:
--------------------------------------------------------------------------------
1 | #include
2 | #include
3 | using namespace std;
4 |
5 | class Solution {
6 | public:
7 | // Member function will return the anwser of string datatype
8 | vector result;
9 | vector generateParenthesis(int n) {
10 | helper("", n, 0, 0);
11 | // Returning the finial answer
12 | return result;
13 | }
14 | void helper(string s, int n, int l, int r){
15 | if (l < r || l > n || r > n) return; // exceed the bundary -> return
16 | if (l == n && r == n){
17 | result.push_back(s);
18 | return;
19 | }
20 | helper(s + "(", n, l+1, r);
21 | helper(s + ")", n, l, r+1);
22 | }
23 | };
24 | //Drive code starts.
25 | int main(){
26 | Solution s;
27 | //calling method generateParentheses() with number.
28 | s.generateParenthesis(4);
29 |
30 | return 0;
31 | }
32 |
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/LeetCode/GenerateParentheses/ques.txt:
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1 | Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
2 |
3 |
4 |
5 | Example 1:
6 |
7 | Input: n = 3
8 | Output: ["((()))","(()())","(())()","()(())","()()()"]
9 | Example 2:
10 |
11 | Input: n = 1
12 | Output: ["()"]
13 |
14 |
15 | Constraints:
16 |
17 | 1 <= n <= 8
18 |
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/LeetCode/Gray Code.cpp:
--------------------------------------------------------------------------------
1 | /*
2 | Problem Statement :
3 |
4 | An n-bit gray code sequence is a sequence of 2n integers where:
5 |
6 | Every integer is in the inclusive range [0, 2n - 1],
7 | The first integer is 0,
8 | An integer appears no more than once in the sequence,
9 | The binary representation of every pair of adjacent integers differs by exactly one bit, and
10 | The binary representation of the first and last integers differs by exactly one bit.
11 | Given an integer n, return any valid n-bit gray code sequence.
12 |
13 | */
14 |
15 |
16 | #include
17 | using namespace std;
18 |
19 | class Solution {
20 | // Recursive function
21 | void recur(int num, int pos, vector& ans) {
22 | ans.push_back(num); // inserting the number in answer vector
23 | if (pos < 0) return; // if position is less than zero then no need to furthur call the function
24 |
25 | // setting the bits of number from pos to zero and calling the function recursively
26 | for (int i = pos; i >= 0; i--)
27 | recur((num | (1 << i)), i - 1, ans); // num | (1 << i) will set the i th bit
28 | }
29 |
30 | public:
31 | vector grayCode(int n) {
32 | vector ans; // defining the answer vector
33 | recur(0, n - 1, ans); // calling the recursive function
34 |
35 | return ans; // return the answer vector
36 | }
37 | };
38 |
39 |
40 | int32_t main() {
41 | Solution solution; // creating a object of solution class
42 |
43 | int n;
44 | cin >> n; // taking the input
45 | vector ans = solution.grayCode(n); // calling the grayCode function of Solution class
46 |
47 | for (auto j : ans)
48 | cout << j << ' ';
49 | cout << endl;
50 |
51 | return 0;
52 | }
53 |
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/LeetCode/House Robber/README.md:
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1 | #### Question link
2 | https://leetcode.com/problems/house-robber/
3 |
--------------------------------------------------------------------------------
/LeetCode/House Robber/question.txt:
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1 | You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
2 |
3 | Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.
4 |
5 |
6 |
7 | Example 1:
8 |
9 | Input: nums = [1,2,3,1]
10 | Output: 4
11 | Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
12 | Total amount you can rob = 1 + 3 = 4.
13 |
14 | Example 2:
15 |
16 | Input: nums = [2,7,9,3,1]
17 | Output: 12
18 | Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
19 | Total amount you can rob = 2 + 9 + 1 = 12.
20 |
21 | Constraints:
22 |
23 | 1 <= nums.length <= 100
24 | 0 <= nums[i] <= 400
25 |
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/LeetCode/Interger To Roman/Question.txt:
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1 | Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
2 |
3 | Symbol Value
4 | I 1
5 | V 5
6 | X 10
7 | L 50
8 | C 100
9 | D 500
10 | M 1000
11 | For example, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.
12 |
13 | Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:
14 |
15 | I can be placed before V (5) and X (10) to make 4 and 9.
16 | X can be placed before L (50) and C (100) to make 40 and 90.
17 | C can be placed before D (500) and M (1000) to make 400 and 900.
18 | Given an integer, convert it to a roman numeral.
19 |
20 |
21 |
22 | Example 1:
23 |
24 | Input: num = 3
25 | Output: "III"
26 | Example 2:
27 |
28 | Input: num = 4
29 | Output: "IV"
30 | Example 3:
31 |
32 | Input: num = 9
33 | Output: "IX"
34 | Example 4:
35 |
36 | Input: num = 58
37 | Output: "LVIII"
38 | Explanation: L = 50, V = 5, III = 3.
39 | Example 5:
40 |
41 | Input: num = 1994
42 | Output: "MCMXCIV"
43 | Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
44 |
45 |
46 | Constraints:
47 |
48 | 1 <= num <= 3999
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/LeetCode/Interger To Roman/Solution.cpp:
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1 | #include
2 | using namespace std;
3 |
4 | class Solution
5 | {
6 | public:
7 | // Member function will return the anwser of string datatype
8 | string intToRoman(int number)
9 | {
10 | // Declaring an integer array of all type of possible numbers to convert it into Roman.
11 | int num[] = {1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000};
12 | // Declearing a string array which will contain all type of possible Romans
13 | string s[] = {"I", "IV", "V", "IX", "X", "XL", "L", "XC", "C", "CD", "D", "CM", "M"};
14 | int i = 12;
15 | string ans = ""; // Taking default answer as an empty string
16 | while (number > 0) // initiates the while loop.
17 | {
18 | int div = number / num[i];
19 | number = number % num[i];
20 | while (div--)
21 | {
22 | ans += s[i];
23 | }
24 | i--;
25 | }
26 | return ans; // Returning the finial answer
27 | }
28 | };
29 |
30 | // { Driver Code Starts.
31 | int main()
32 | {
33 | //calling method intToRoman() with number.
34 | Solution s;
35 | s.intToRoman(3);
36 |
37 | return 0;
38 | }
39 |
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/LeetCode/Interger To Roman/readme.md:
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1 | Link to the question - https://leetcode.com/problems/integer-to-roman/
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/LeetCode/Island perimeter/question.txt:
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1 | You are given row x col grid representing a map where grid[i][j] = 1 represents land and grid[i][j] = 0 represents water.
2 |
3 | Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells).
4 |
5 | The island doesn't have "lakes", meaning the water inside isn't connected to the water around the island. One cell is a square with side length 1. The grid is rectangular, width and height don't exceed 100. Determine the perimeter of the island.
6 |
7 | Input: grid = [[0,1,0,0],[1,1,1,0],[0,1,0,0],[1,1,0,0]]
8 | Output: 16
9 | Explanation: The perimeter is the 16 yellow stripes in the image above.
10 | Example 2:
11 |
12 | Input: grid = [[1]]
13 | Output: 4
14 | Example 3:
15 |
16 | Input: grid = [[1,0]]
17 | Output: 4
18 |
19 |
20 | Constraints:
21 |
22 | row == grid.length
23 | col == grid[i].length
24 | 1 <= row, col <= 100
25 | grid[i][j] is 0 or 1.
26 | There is exactly one island in grid.
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/LeetCode/Island perimeter/readme.md:
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1 | question link: https://leetcode.com/problems/island-perimeter/
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/LeetCode/Island perimeter/solution.cpp:
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1 | //https://leetcode.com/problems/island-perimeter/
2 |
3 | class Solution {
4 | public:
5 | int islandPerimeter(vector>& grid) {
6 | int result=0;
7 | int row = grid.size(), col = grid[0].size(); // row and column
8 | if(row==0 || col==0)
9 | return 0;
10 | for(int i=0;i0 && grid[i][j-1]==1) // chek if has left block
19 | {
20 | result=result-2;
21 | }
22 |
23 | if(i>0 && grid[i-1][j]==1) // check if it has up block
24 | {
25 | result=result-2;
26 | }
27 |
28 | }
29 | }
30 | }
31 | return result;
32 | }
33 | };
34 |
35 |
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/LeetCode/Jump Game II/Question.txt:
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1 | Given an array of non-negative integers nums, you are initially positioned at the first index of the array.
2 |
3 | Each element in the array represents your maximum jump length at that position.
4 |
5 | Your goal is to reach the last index in the minimum number of jumps.
6 |
7 | You can assume that you can always reach the last index.
8 |
9 |
10 |
11 | Example 1:
12 |
13 | Input: nums = [2,3,1,1,4]
14 | Output: 2
15 | Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
16 | Example 2:
17 |
18 | Input: nums = [2,3,0,1,4]
19 | Output: 2
20 |
21 |
22 | Constraints:
23 |
24 | 1 <= nums.length <= 104
25 | 0 <= nums[i] <= 1000
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/LeetCode/Jump Game II/readme.md:
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1 | Question link : https://leetcode.com/problems/jump-game-ii/
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/LeetCode/Jump Game II/solution.cpp:
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1 |
2 | //https://leetcode.com/problems/jump-game-ii/
3 |
4 |
5 | // Time complexity O(N)
6 | // Auxillary space O(1)
7 |
8 | /*
9 | Approach:
10 | 1.maxReach means the maximum we can reach from that particular index
11 | which is the index plus the value of the index (the jump value).
12 | So, we keep updating it in each iteration so that whenever we move forward,
13 | the variable stores the maximum we can reach by using maxReach = max(maxReach, nums[i]+i)
14 | 2.Also, at each iteration, we reduce our steps variable by 1.
15 | 3.So, in this solution, we return jumps+1 as our answer.
16 | */
17 |
18 | class Solution {
19 | public:
20 | int jump(vector& nums) {
21 | if(nums.size()==1)
22 | return 0;
23 | int maxReach = nums[0];
24 | int steps = nums[0];
25 | int jumps = 0;
26 | for(int i=1;i& nums) {
4 | int n= nums.size(); //size of given vector
5 |
6 | bool dp[n]; //create a dp table of size equal n and initialize all values as false
7 | for(int i=0;i letterCombinations(string digits) {
29 | vectorphone={"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"}; //create vector of string for each number in phone
30 | if(digits.empty())
31 | return {}; //if input is empty return empty vector
32 | vectorresult; //store final result
33 | result.push_back("");
34 | for(auto d:digits){ //for every digit
35 | vectortemp; //create temp vecotr to store string cur string and update in result vector
36 | for(auto s:phone[d-'0']){
37 | for(auto r:result){
38 | temp.push_back(r+s); //compute every combination of strings
39 | }
40 | }
41 | result.swap(temp);
42 | }
43 | return result;
44 | }
45 | };
46 |
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/LeetCode/Merge k Sorted Lists/question.txt:
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1 |
2 | You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.
3 |
4 | Merge all the linked-lists into one sorted linked-list and return it.
5 |
6 |
7 |
8 | Example 1:
9 |
10 | Input: lists = [[1,4,5],[1,3,4],[2,6]]
11 | Output: [1,1,2,3,4,4,5,6]
12 | Explanation: The linked-lists are:
13 | [
14 | 1->4->5,
15 | 1->3->4,
16 | 2->6
17 | ]
18 | merging them into one sorted list:
19 | 1->1->2->3->4->4->5->6
20 | Example 2:
21 |
22 | Input: lists = []
23 | Output: []
24 | Example 3:
25 |
26 | Input: lists = [[]]
27 | Output: []
28 |
29 |
30 | Constraints:
31 |
32 | k == lists.length
33 | 0 <= k <= 10^4
34 | 0 <= lists[i].length <= 500
35 | -10^4 <= lists[i][j] <= 10^4
36 | lists[i] is sorted in ascending order.
37 | The sum of lists[i].length won't exceed 10^4.
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/LeetCode/Merge k Sorted Lists/readme.md:
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1 | https://leetcode.com/problems/merge-k-sorted-lists/
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/LeetCode/MiddleLL.cpp:
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1 | //https://leetcode.com/problems/middle-of-the-linked-list/
2 |
3 | #include
4 | using namespace std;
5 | class Node{//creating node
6 | public:
7 | int data;
8 | Node *next;
9 | };
10 | class NodeOperation{
11 | public:
12 | void pushNode(class Node** head_ref,int data_val){//adding elements to linked list
13 | class Node *new_node = new Node();
14 | new_node->data = data_val;
15 | new_node->next = *head_ref;
16 | *head_ref = new_node;
17 | }
18 |
19 | void printNode(class Node *head){//printing linked list
20 | while(head != NULL){
21 | cout <data << "->";
22 | head = head->next;
23 | }
24 | cout << "NULL" << endl;
25 | }
26 |
27 | void printMiddle(class Node *head){//code to print the middle of linked list
28 | struct Node *slow_ptr = head;
29 | struct Node *fast_ptr = head;
30 | if (head!=NULL)
31 | {
32 | while (fast_ptr != NULL && fast_ptr->next != NULL)
33 | {
34 | fast_ptr = fast_ptr->next->next;
35 | slow_ptr = slow_ptr->next;
36 | }
37 | cout << "The middle element is [" << slow_ptr->data << "]" << endl;
38 | }
39 | }
40 | };
41 |
42 | int main(){//main function
43 | class Node *head = NULL;
44 | class NodeOperation *temp = new NodeOperation();
45 | for(int i=5; i>0; i--){//control loop to push elements into array
46 | temp->pushNode(&head, i);
47 | temp->printNode(head);
48 | temp->printMiddle(head);
49 | }
50 | return 0;
51 | }
52 |
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/LeetCode/Permutation.cpp:
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1 | //Problem link:https://leetcode.com/problems/permutations/
2 | class Solution {
3 | public:
4 | vector> result;
5 |
6 | // Backtracking
7 | void permutation(vector &nums,int i,int n){
8 | if(i==n){
9 | result.push_back(nums);
10 | return ;
11 | }
12 |
13 | for(int j=i;j<=n;j++){
14 | swap( nums[i],nums[j]);
15 | permutation(nums,i+1,n); //index is increased by one so that swapping is done with next index element in further calls
16 | swap( nums[i],nums[j]); //swapping it back to original order after recursion call is over
17 | }
18 | }
19 |
20 | vector> permute(vector& nums) {
21 |
22 | permutation(nums,0,nums.size()-1);
23 |
24 | return result;
25 | }
26 | };
27 |
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/LeetCode/Power-of-four/Readme.md:
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1 | Question Link- https://leetcode.com/problems/power-of-four/
2 |
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/LeetCode/Power-of-four/Solution.cpp:
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1 | //Given an integer n, return true if it is a power of four. Otherwise, return false.
2 | #include
3 | using namespace std;
4 | #define ll long long
5 | bool isPowerOfFour(int n) {
6 |
7 | if(n<=0)
8 | {
9 | return false; //Negative Numbers can't be a power of 4
10 | }
11 | return !(n&(n-1)) and n%3==1; //If a number is power of 2 then n7(n-1) is zero and if the number is power of 4 it always gives 1 when taken mod with 3
12 |
13 |
14 | }
15 | int main() {
16 | int n;
17 | cin>>n; //Taking Input the number
18 | bool ans=isPowerOfFour(n);
19 | if(ans)
20 | {
21 | cout<<"Yes "< findRepeatedDnaSequences(string s) {
30 | mapstore;
31 | vectorans;
32 | int i=0,n=s.length();
33 | while(i+10<=n){
34 | string sub=s.substr(i++,10); //extracting all substring of length from string
35 | store[sub]++; //increment count if same string occurs
36 | if(store[sub]==2) //check if it is repeated again if so then add it to vector
37 | ans.push_back(sub);
38 | }
39 | return ans; //return final vector
40 | }
41 | };
42 |
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/LeetCode/ReverseInteger/ques.txt:
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1 | Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-2^31, 2^31 - 1], then return 0.
2 |
3 | Assume the environment does not allow you to store 64-bit integers (signed or unsigned).
4 |
5 |
6 |
7 | Example 1:
8 |
9 | Input: x = 123
10 | Output: 321
11 | Example 2:
12 |
13 | Input: x = -123
14 | Output: -321
15 | Example 3:
16 |
17 | Input: x = 120
18 | Output: 21
19 | Example 4:
20 |
21 | Input: x = 0
22 | Output: 0
23 |
24 |
25 | Constraints:
26 |
27 | -2^31 <= x <= 2^31 - 1
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/LeetCode/ReverseInteger/reverseInteger.cpp:
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1 | #include
2 | using namespace std;
3 |
4 | class Solution{
5 | public:
6 | // Member function will return the anwser of int datatype
7 | int reverse(int x){
8 | int rev=0;
9 | while(x!=0){
10 | int digit=x%10;
11 | x/=10;
12 |
13 | if (rev > 2147483647/10){
14 | return 0;
15 | }
16 | if (rev < -2147483647/10){
17 | return 0;
18 | }rev=rev*10+digit;
19 | // Returning the finial answer
20 | }return rev;
21 | }
22 | };
23 | // { Driver Code Starts.
24 | int main(){
25 | Solution s;
26 | //calling method reverseI() with number.
27 | s.reverse(4);
28 |
29 | return 0;
30 | }
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/LeetCode/Search Insert Position.cpp:
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1 | //https://leetcode.com/problems/search-insert-position/
2 | class Solution {
3 | public:
4 | int searchInsert(vector& nums, int target) {
5 | //Apply Binary search on array for target value and for position return low value;
6 | int low=0;
7 | int high=nums.size()-1;
8 | while(low<=high){
9 | int mid=(low+high)/2;
10 | if(nums[mid]>target)
11 | high=mid-1;
12 | else if(nums[mid]=0;i--){
7 | if(k>25){ //check balance k greater than 25 if it is then add 'z' at end
8 | s[i]='z';
9 | k=k-25;
10 | }
11 | else{ //else we add the character if it is less than equal 25
12 | s[i]=97+k;
13 | break;
14 | }
15 | }
16 | return s;
17 | }
18 | };
19 |
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/LeetCode/Sorting Ascending.cpp:
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1 | //sort the colors input by the user. take red as 0, white as 1 and blue as 2
2 | #include
3 | using namespace std;
4 | int main()
5 | {
6 | cout<<"No. of entries: \n";
7 | int n;
8 | cin>>n;
9 | int a[n],t;//an array with size n and a temporary variable that will later be used for swapping
10 | cout<<"Enter "< ";
14 | cin>>a[i];
15 | cout<<"\n";
16 | }// shows output as: "->0
17 | // ->2
18 | // ->1 and so on
19 | for(int i=0;i0){
7 | int digit=x%10;
8 | sum+=digit;
9 | mul*=digit;
10 | x=x*10;
11 | }
12 | return mul-sum;
13 | }
14 | };
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/LeetCode/Welcome.txt:
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https://raw.githubusercontent.com/SarthakKeshari/CPP-Questions-and-Solutions/b3097685322caac5124ecfbfd41a221fb6bc8318/LeetCode/Welcome.txt
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/LeetCode/ZigZagConversion/ques.txt:
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1 | The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
2 |
3 | P A H N
4 | A P L S I I G
5 | Y I R
6 | And then read line by line: "PAHNAPLSIIGYIR"
7 |
8 | Write the code that will take a string and make this conversion given a number of rows:
9 |
10 | string convert(string s, int numRows);
11 |
12 |
13 | Example 1:
14 |
15 | Input: s = "PAYPALISHIRING", numRows = 3
16 | Output: "PAHNAPLSIIGYIR"
17 | Example 2:
18 |
19 | Input: s = "PAYPALISHIRING", numRows = 4
20 | Output: "PINALSIGYAHRPI"
21 | Explanation:
22 | P I N
23 | A L S I G
24 | Y A H R
25 | P I
26 | Example 3:
27 |
28 | Input: s = "A", numRows = 1
29 | Output: "A"
30 |
31 |
32 | Constraints:
33 |
34 | 1 <= s.length <= 1000
35 | s consists of English letters (lower-case and upper-case), ',' and '.'.
36 | 1 <= numRows <= 1000
--------------------------------------------------------------------------------
/LeetCode/trappingRainWater.cpp:
--------------------------------------------------------------------------------
1 |
2 | //https://leetcode.com/problems/trapping-rain-water/
3 |
4 | class Solution {
5 | public:
6 | int trap(vector& height) {
7 |
8 | int n = height.size();
9 | int total=0;
10 | int left = 0, right = n-1; //Taking first element as left and last element as right
11 |
12 | int maxLeft=0, maxRight=0; //Consider max height for left and right respectively
13 |
14 | while(left <= right){ //Check till left element is small or equal to right
15 |
16 | if(height[left]<=height[right]){ //check till height for left is less or equal to right
17 |
18 | if(height[left] >= maxLeft) //Height at Left is greater or equal to maximum value at left
19 | maxLeft = height[left];
20 | else
21 | total += maxLeft - height[left]; //Add the difference between maximum value and height at left
22 |
23 | left++; //Increment from left side
24 | }
25 | else{
26 | if(height[right] >= maxRight){ //Height at Right is greater or equal to maximum value at right
27 | maxRight = height[right];
28 | }
29 | else
30 | total += maxRight - height[right]; //Add the difference between maximum value and height at right
31 |
32 | right--; //Decrement from right side
33 | }
34 | }
35 | return total; //Return the total units of water trapped between bars
36 | }
37 | };
38 |
--------------------------------------------------------------------------------
/Number Theory/Special_Number.cpp:
--------------------------------------------------------------------------------
1 | #include
2 | using namespace std;
3 |
4 | int factorial(int n){
5 | if (n==0)
6 | return 1;
7 | return (n*factorial(n-1));
8 | }
9 |
10 | int sum(int num){
11 | int s=0;
12 | while(num>0){
13 | s=s+factorial(num%10);
14 | num=num/10;
15 | }
16 | return s;
17 | }
18 |
19 | int main(){
20 | int number,add;
21 | cin>>number;
22 | add=sum(number);
23 | if(add==number)
24 | cout<
13 |
14 | using namespace std;
15 |
16 | int main()
17 | {
18 | int n=5;
19 | for(int i=1;i<=n;i++)
20 | {
21 | for(int j=1;j<=i;j++)
22 | {
23 | cout<<"*";
24 | }
25 | cout< //class for handling input and output
3 |
4 | using namespace std;
5 |
6 | int main()
7 | {
8 | int x, y; //temporary variables to store the row and column number
9 | for(x=0 ; x<=8 ; x++) //to change row
10 | {
11 | for(y=0 ; y<=11 ; y++) //to change column
12 | {
13 | if((y!=5&&y!=6)&&x==0)
14 | {
15 | cout << "$"; //printing statement
16 | }
17 | else if(((y==5||y==6)&&x!=8)||(x==6&&(y==0||y==1))||(x==7&&(y==1||y==2))||(x==8&&(y==3||y==4)))
18 | {
19 | cout << "#"; //printing statement
20 | }
21 | else
22 | {
23 | cout << " "; //printing statement for space
24 | }
25 | }
26 | cout << "\n"; //printing statement for next line
27 | }
28 | return 0;
29 | }
30 |
--------------------------------------------------------------------------------
/Pattern printing/Letter W.cpp:
--------------------------------------------------------------------------------
1 | //Write a program to print letter W exactly as shown
2 | #include //class responsible for handling input and output
3 |
4 | using namespace std;
5 |
6 | int main()
7 | {
8 | int x,y; //temporary integer type variables to store row and column number
9 | for(x=0 ; x<=5 ; x++) //To change the rows
10 | {
11 | for(y=0 ; y<=22 ; y++) //To change the columns
12 | {
13 | if((x==y)||(y==(x+1))||((22-y)==x)||((22-y)==(x+1)))
14 | {
15 | cout << "@"; //printing statement
16 | }
17 | else if(((x+y)==11)||((y-x)==11))
18 | {
19 | cout << "#"; //printing statement
20 | }
21 | else
22 | {
23 | cout << " "; //printing statement for space
24 | }
25 | }
26 | cout << "\n"; //printing statement for next line
27 | }
28 | return 0;
29 | }
30 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_A.cpp:
--------------------------------------------------------------------------------
1 | /*
2 | This code prints the letter A, as shown below.
3 |
4 | **
5 | ** **
6 | ** **
7 | ** **
8 | ** **
9 | **########**
10 | ** **
11 | ** **
12 | ** **
13 |
14 | There are 6 distinct patterns featured in printing the letter A to console.
15 | If we define those patterns as strings, and also define the order in which
16 | they are printed, then we could iterate over the order and print the correct
17 | pattern each time, resulting in the correct output. (i.e., the letter A)
18 | */
19 | #include
20 |
21 | int main()
22 | {
23 | std::string essentialStrings[6] = {" **", " ** **", " ** **", " ** **", "** **", "**########**"};
24 | int n = 9;
25 | int printIndices[9] = {0, 1, 2, 3, 4, 5, 4, 4, 4};
26 | for(int i=0; i < n; ++i)
27 | {
28 | std::cout << essentialStrings[ printIndices[i] ] << std::endl;
29 | }
30 |
31 | return 0;
32 | }
33 |
34 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_B.cpp:
--------------------------------------------------------------------------------
1 | /*
2 | This code prints the letter B, as shown below.
3 |
4 | **#####
5 | ** ##
6 | ** ##
7 | ** ##
8 | **#####
9 | ** ##
10 | ** ##
11 | ** ##
12 | **#####
13 |
14 | There are 3 distinct patterns featured in printing the letter B to console.
15 | If we define those patterns as strings, and also define the order in which
16 | they are printed, then we could iterate over the order and print the correct
17 | pattern, resulting in the correct output. (i.e., the letter B)
18 | */
19 | #include
20 |
21 | int main()
22 | {
23 | std::string essentialStrings[3] = {"**#####", "** ##", "** ##"};
24 | int n = 9;
25 | int printIndices[9] = {0, 1, 2, 1, 0, 1, 2, 1, 0};
26 | for(int i=0; i < n; ++i)
27 | {
28 | std::cout << essentialStrings[ printIndices[i] ] << std::endl;
29 | }
30 |
31 | return 0;
32 | }
33 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_C.cpp:
--------------------------------------------------------------------------------
1 | /*
2 | This code prints the letter C, as shown below.
3 |
4 | #####
5 | ** **
6 | ** **
7 | ##
8 | ##
9 | ##
10 | ##
11 | ##
12 | ** **
13 | ** **
14 | #####
15 |
16 | There are 4 distinct patterns featured in printing the letter C to console.
17 | If we define those patterns as strings, and also define the order in which
18 | they are printed, then we could iterate over the order and print the correct
19 | overall pattern, resulting in the correct output. (i.e., the letter C)
20 | */
21 | #include
22 |
23 | int main()
24 | {
25 | std::string essentialStrings[4] = {" #####", " ** **", " ** **", "##"};
26 | int n = 11;
27 | int printIndices[11] = {0, 1, 2, 3, 3, 3, 3, 3, 2, 1, 0};
28 | for(int i=0; i < n; ++i)
29 | {
30 | std::cout << essentialStrings[ printIndices[i] ] << std::endl;
31 | }
32 |
33 | return 0;
34 | }
35 |
36 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_E.cpp:
--------------------------------------------------------------------------------
1 | // C++ program to print the below pattern
2 |
3 | // ##$$$$$$$
4 | // ##
5 | // ##
6 | // ##
7 | // ##$$$$$$$
8 | // ##
9 | // ##
10 | // ##
11 | // ##$$$$$$$
12 |
13 | #include
14 | using namespace std;
15 |
16 | int main() {
17 | // Printing the pattern
18 |
19 | // As the pattern is of 9x9 size, therefore a nested for-loop from 0-8 is peformed.
20 |
21 | for (int i = 0; i < 9; i++)
22 | {
23 | for (int j = 0; j < 9; j++)
24 | {
25 | // To print the first two # columns.
26 | if (j < 2)
27 | {
28 | cout << "#";
29 | }
30 |
31 | // To print the $ rows.
32 | else if ((i == 0 or i == 4 or i == 8) and j >= 2)
33 | {
34 | cout << "$";
35 | }
36 |
37 | // To print the empty spaces.
38 | else
39 | {
40 | cout << " ";
41 | }
42 | }
43 |
44 | cout << endl;
45 | }
46 |
47 | return 0;
48 | }
49 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_F.cpp:
--------------------------------------------------------------------------------
1 | // C++ program to print the below pattern
2 |
3 | // ##$$$$$$$
4 | // ##
5 | // ##
6 | // ##
7 | // ##$$$$$$$
8 | // ##
9 | // ##
10 | // ##
11 | // ##
12 |
13 | #include
14 | using namespace std;
15 |
16 | int main() {
17 | // Printing the pattern
18 |
19 | // As the pattern is of 9x9 size, therefore a nested for-loop from 0-8 is peformed.
20 | for (int i = 0; i < 9; i++)
21 | {
22 | for (int j = 0; j < 9; j++)
23 | {
24 | // To print the first two # columns.
25 | if (j < 2)
26 | {
27 | cout << "#";
28 | }
29 |
30 | // To print the $ rows.
31 | else if ((i == 0 or i == 4) and j >= 2)
32 | {
33 | cout << "$";
34 | }
35 |
36 | // To print the empty spaces.
37 | else
38 | {
39 | cout << " ";
40 | }
41 | }
42 |
43 | cout << endl;
44 | }
45 |
46 | return 0;
47 | }
48 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_H.cpp:
--------------------------------------------------------------------------------
1 | // C++ program to print the below pattern
2 | /*
3 | ## ##
4 | ## ##
5 | ## ##
6 | ## ##
7 | ##$$$$$##
8 | ## ##
9 | ## ##
10 | ## ##
11 | ## ##
12 | */
13 | #include
14 | using namespace std;
15 | int main(){
16 | //For printing the 9*9 letter H.
17 | for(int i = 0;i<9;i++){
18 | for(int j = 0;j<9;j++){
19 | //For printing the First two and last two columns.
20 | if(j<2 ||(j>=7 && j<9)){
21 | cout << '#';
22 | }else{
23 | if(i==4){
24 | //For printing the Middle Lyer of $ sign
25 | cout << '$';
26 | }else{
27 | cout << ' ';
28 | }
29 | }
30 | }
31 | cout << '\n';
32 | }
33 | return 0;
34 | }
35 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_I.cpp:
--------------------------------------------------------------------------------
1 | //C++ program to print the below pattern :-
2 |
3 | /* $$$$##$$$$
4 | ##
5 | ##
6 | ##
7 | ##
8 | ##
9 | ##
10 | ##
11 | $$$$##$$$$ */
12 |
13 | #include
14 | #define ll long long
15 | #define N 100005
16 | #define mod 1000000007
17 | #define INF 2147483647
18 | #define ln "\n"
19 | #define fast() ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);
20 | using namespace std;
21 | int main()
22 | {
23 | fast();
24 |
25 | // for printing the pattern I have used 3 distinct patterns for printing the letter I
26 |
27 | string S1[3] = {"$$$$##$$$$", " ##", "$$$$##$$$$"};
28 | int number = 9;
29 | int print[9] = {0,1, 1, 1, 1, 1, 1, 1, 0}; //iterating over the correct sequence and printing the correct pattern
30 | for(int j=0; j < number ;j++) // for loop from 0 to n
31 | {
32 | cout << S1[ print[j] ] << ln; // prints the final pattern
33 | }
34 | return 0;
35 | }
36 |
37 | // This Code is contributed by Adarsh Kumar Singh (Username ----> Fury-MadMax)
38 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_K.cpp:
--------------------------------------------------------------------------------
1 | //Question:
2 | /* Write a CPP Program to print the below pattern:
3 |
4 | ## @@
5 | ## @@
6 | ## @@
7 | ## @@
8 | ##@@
9 | ## @@
10 | ## @@
11 | ## @@
12 | ## @@
13 |
14 | */
15 | #include
16 | using namespace std;
17 |
18 |
19 | int main(){
20 | ios_base::sync_with_stdio(false);
21 | cin.tie(NULL);
22 |
23 | int n;//there are a total of n rows in the pattern
24 | cin>>n;//taking input as number of rows in the letter
25 | //loop for the first/upper half of the letter K
26 | for (int i = 0; i < n-n/2; i++)
27 | {
28 | cout<<"##";//printing first ## in every row
29 | for (int j = 0; j < n-n/2-i-1; j++)
30 | {
31 | cout<<" ";//printing spaces whihc decrease in the upper half
32 | }
33 | cout<<"@@"< is the "##" part - printed in a straight line for 7 times, for this we run a for loop ,
16 | second-> is the "##&&&&&&&" part - this can be printed directly as printf("##&&&&&&&\n").
17 |
18 | Hence we get the complete pattern.
19 |
20 | */
21 |
22 | #include
23 | using namespace std;
24 | int main()
25 | {
26 | //To print ##
27 | for(int i=0; i < 7; ++i) //printing ## in this loop for 7 times (n = 0 to 7)
28 | {
29 | printf("##\n"); // in a straight line
30 |
31 | }
32 | printf("##&&&&&&&\n"); //print one line "##&&&&&&&" to complete the last line of the letter L
33 |
34 | return 0; //we get the letter L pattern with the charecters needed.
35 | }
36 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_M.cpp:
--------------------------------------------------------------------------------
1 | /*
2 | CPP Program to print Letter M
3 |
4 | ##@@ @@##
5 | ## @@ @@ ##
6 | ## @@ @@ ##
7 | ## ## ##
8 | ## ##
9 | ## ##
10 | ## ##
11 | ## ##
12 |
13 |
14 | Letter M can be broken down into 3 parts
15 | - full length first ## column
16 | - half length V letter
17 | - full length last ## column
18 | */
19 | #include
20 |
21 | using namespace std;
22 |
23 | int main() {
24 | int level = 8;
25 | int vLevel = level/2;
26 |
27 | int firstSpaces = 0, lastSpaces = 0, midSpaces = 0;
28 | int i, cols = 2 * vLevel + 2;
29 |
30 | //for printing pattern till last row
31 | for (i = 0; i < level; ++i) {
32 | cout << "##";
33 | if (i < vLevel -1) {
34 | firstSpaces = i;
35 | lastSpaces = i;
36 | midSpaces = cols - firstSpaces - lastSpaces - 4; //for @@ and @@
37 |
38 | //print leading spaces
39 | cout << string(firstSpaces, ' ');
40 | //print @@
41 | cout << "@@";
42 |
43 | //print mid spaces
44 | cout << string(midSpaces, ' ');
45 |
46 | //print @@
47 | cout << "@@";
48 |
49 | //print last spaces;
50 | cout << string(lastSpaces, ' ');
51 | }
52 | else if (i == vLevel -1) { //last level of V
53 | cout << string(i + 1, ' ') << "##" << string(i + 1, ' ');
54 | }
55 | else {
56 | cout << string(cols, ' ');
57 | }
58 | cout << "##";
59 | cout << endl;
60 | }
61 | return 0;
62 | }
63 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_N.cpp:
--------------------------------------------------------------------------------
1 | /*
2 |
3 | This code prints the letter N, as shown below.
4 |
5 | ##@@ ##
6 | ## @@ ##
7 | ## @@ ##
8 | ## @@ ##
9 | ## @@ ##
10 | ## @@ ##
11 | ## @@ ##
12 | ## @@##
13 |
14 | LOGIC:
15 | We need to run nested loops here , like a matrix i for depth - 8, j for width - 8
16 | First is the straight line "##",
17 | that is covered by for (i = 0; i < 8; i++).
18 | for i = 0 -
19 | we get ##@@ ##
20 | i = 1 ## @@ ##
21 | i = 2 ## @@ ##
22 | .
23 | .
24 | .
25 | and so on
26 | counter increments on every completed cycle,
27 | j handles the pattern on the horizontal i.e. printing "@@" and the second line of "##".
28 | Hence we get the desired Pattern N.
29 |
30 | */
31 |
32 | #include
33 | using namespace std;
34 |
35 | // function to print the letter N Pattern
36 | void printN()
37 |
38 | {
39 | int i, j, counter = 0;
40 |
41 | //to print ## and @@
42 | for (i = 0; i < 8; i++) // taking i<8 to print "##" in a straight line, 7 times (n = 0 to 8).
43 | {
44 | cout <<"##"; //this prints "##" in a striaght line .
45 |
46 | // to print "@@".
47 | for (j = 0; j <= 8; j++)
48 | {
49 | if (j == 8) // to print the second line of "##"
50 | cout <<"##";
51 | else if (j == counter)
52 | cout <<"@@";
53 | else
54 | cout <<" "; //else print " " according to the pattern N required.
55 | }
56 | counter++;
57 | cout <<"\n";
58 | }
59 | }
60 |
61 | int main(int argc, char** argv) //To print the function N
62 | {
63 | printN(); // Letter N as output
64 | }
65 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_V.cpp:
--------------------------------------------------------------------------------
1 | /*
2 | CPP Program to print Letter V
3 |
4 |
5 | @@ @@
6 | @@ @@
7 | @@ @@
8 | @@ @@
9 | @@ @@
10 | @@ @@
11 | @@@
12 | */
13 |
14 | #include
15 |
16 | using namespace std;
17 |
18 | int main() {
19 | int level = 7;
20 |
21 | int firstSpaces = 0, lastSpaces = 0, midSpaces = 0;
22 | int i, cols = 2 * level + 1;
23 |
24 | //for printing pattern till last row
25 | for (i = 0; i < level - 1; ++i) {
26 | firstSpaces = i;
27 | lastSpaces = i;
28 | midSpaces = cols - firstSpaces - lastSpaces - 4; //for @@ and @@
29 |
30 | //print leading spaces
31 | cout << string(firstSpaces, ' ');
32 | //print @@
33 | cout << "@@";
34 |
35 | //print mid spaces
36 | cout << string(midSpaces, ' ');
37 | //print @@
38 | cout << "@@";
39 | //print last spaces;
40 | cout << string(lastSpaces, ' ');
41 | cout << endl;
42 | }
43 | //print last line
44 | cout << string(level - 1, ' ') << "@@@" << string(level - 1, ' ') << endl;
45 |
46 | return 0;
47 | }
48 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_X.cpp:
--------------------------------------------------------------------------------
1 | //C++ program to print the below pattern :-
2 |
3 | /*
4 | @@ ##
5 | @@ ##
6 | @@ ##
7 | @@ ##
8 | @@@
9 | ## @@
10 | ## @@
11 | ## @@
12 | ## @@
13 | */
14 |
15 | #include
16 | using namespace std;
17 |
18 | int main()
19 | {
20 | int row,coloumn;
21 |
22 | for(row=0; row<9; row++) // here 9 is the total no. of rows.
23 | {
24 | for(coloumn=0; coloumn<9; coloumn++) // here 9 is the total no. of coloumns.
25 | {
26 | if(row==4 && coloumn==4)
27 | cout << "@@@"; // printing @@@ in the middle of the letter when rows=4 & coloumns=4
28 |
29 | else if(row==coloumn) // prints the required characters when number of rows equals number of columns
30 | cout << "@@";
31 |
32 | else if((row+coloumn)==8) // prints the required characters when number of rows + number of columns equals (total rows-1)
33 | cout << "##";
34 |
35 | else
36 | cout << " "; // prints a whitespace at places where characters are not printed.
37 | }
38 | cout << "\n"; // prints a newline at the end of each row.
39 | }
40 | return 0;
41 | }
42 |
43 |
44 | // This Code is contributed by Harshvardhan Singh (Username ----> cyberskull99297)
45 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_Y.cpp:
--------------------------------------------------------------------------------
1 | //C++ program to print the below pattern :-
2 |
3 | /*
4 | @@ ##
5 | @@ ##
6 | @@ ##
7 | @@ ##
8 | @@##
9 | @@
10 | @@
11 | @@
12 | @@
13 | */
14 |
15 | #include
16 | using namespace std;
17 |
18 | int main()
19 | {
20 | int row,column;
21 |
22 | for(row=0; row<9; row++) // here 9 is the total no. of rows.
23 | {
24 | for(column=0; column<=9; column++) // here 9 is the total no. of columns.
25 | {
26 | if(row==4 && column==4) // prints '@@' when rows=4 & column=4
27 | cout << "@@";
28 |
29 | else if(row==column && row<5) // prints the required character when the number of row is less than 5.
30 | cout << "@@";
31 |
32 | else if((row+column)==9 && row<5) // prints the required character when the number of row is less than 5.
33 | cout << "##";
34 |
35 | else if(column==5 && row>=5) // prints the required character when the number of row is greater than or equal to 5.
36 | cout << "@@";
37 |
38 | else
39 | cout << " "; // prints a whitespace at places where characters are not printed.
40 |
41 | }
42 | cout << "\n"; // prints a newline at the end of each row.
43 | }
44 |
45 | return 0;
46 | }
47 |
48 | // This Code is contributed by Harshvardhan Singh (Username ----> cyberskull99297)
49 |
--------------------------------------------------------------------------------
/Pattern printing/Letter_Z.cpp:
--------------------------------------------------------------------------------
1 | /*
2 |
3 | This code will print pattern like this......
4 |
5 | @@@@@@@@@
6 | ##
7 | ##
8 | ##
9 | ##
10 | ##
11 | ##
12 | @@@@@@@@@
13 |
14 | */
15 |
16 | #include
17 | using namespace std;
18 |
19 | int main() {
20 | // your code goes here
21 | int x,y;
22 | // For printing 8*9 letter Z
23 | for(x=0;x<8;x++){
24 | for(y=0;y<9;y++){
25 |
26 | if(x==0 || x==7){
27 | // For printing top and bottom horizontal line of letter Z using '@'
28 | cout<<"@";
29 | }
30 | else{
31 | // For printing slant line of letter Z using '#'
32 | if(y==(9-x-2)){
33 | cout<<"##";
34 | y++;
35 | }
36 | else{
37 | cout<<" ";
38 | }
39 | }
40 | }
41 | cout<<"\n";
42 | }
43 | return 0;
44 | }
45 |
--------------------------------------------------------------------------------
/Pattern printing/butterfly Pattern.cpp:
--------------------------------------------------------------------------------
1 | // Make a butterfly pattern
2 |
3 | #include
4 | using namespace std;
5 |
6 | int main()
7 | {
8 | int n;
9 | cout<<"Enter the value of n: ";
10 | cin>>n; // taken a input from user
11 | for(int i=1;i<=n;i++){ //used for loop for *
12 | for(int j=1;j<=i;j++){
13 | cout<<"*"; //printed *
14 | }
15 | int space=2*n-2*i;
16 | for(int j=1;j<=space;j++){ // used for loop for Space
17 | cout<<" "; //printed space
18 | }
19 | for(int j=1;j<=i;j++){ //again used for loop for *
20 | cout<<"*"; // printed *
21 | }
22 | cout<=1;i--){ //used for loop for *
25 | for(int j=i;j>=1;j--){
26 | cout<<"*"; //printed *
27 | }
28 | int space=2*n-2*i;
29 | for(int j=space;j>=1;j--){ // used for loop for Space
30 | cout<<" "; //printed space
31 | }
32 | for(int j=i;j>=1;j--){ //again used for loop for *
33 | cout<<"*"; //printed *
34 | }
35 | cout<
3 | using namespace std;
4 | // Recursive function to return gcd of a and b
5 | int gcd(int a, int b){
6 | if (a == 0)
7 | return b;
8 | if (b == 0)
9 | return a;
10 | if (a == b) //base case of recursion
11 | return a;
12 | if (a > b)
13 | return gcd(a-b, b);
14 | return gcd(a, b-a);
15 | }
16 |
17 | int main(){
18 | int a = 98, b = 56;
19 | cout<<"GCD of "<
5 | using namespace std;
6 |
7 | // Function to print first n rows of Pascal's triangle
8 | void printPascal(int Row_num)
9 | {
10 | int arr[Row_num][Row_num];
11 |
12 | for (int i = 0; i < Row_num; i++)
13 | {
14 | for (int j = 0; j <= i; j++)
15 | {
16 | if (i == j || j == 0) //Making first and last element of Every row as 1
17 | arr[i][j] = 1;
18 | else
19 | arr[i][j] = arr[i-1][j - 1] + arr[i- 1][j]; //Adding the element of pascal triangle by two corresponding elements
20 | cout << arr[i][j] << " ";
21 | }
22 | cout <>n;
46 | cout<
2 | using namespace std;
3 |
4 | int factorial(int i){
5 | if(i==1)
6 | return 1;
7 |
8 | return i*factorial(i-1);
9 |
10 | }
11 |
12 | int main(){
13 | int number;
14 | cout<<"Enter number : ";
15 | cin>>number;
16 | cout<
16 | using namespace std;
17 |
18 | int main()
19 | {
20 | // number of elements
21 | int n;
22 | cin>>n;
23 |
24 | //take input of array elements
25 | int arr[n];
26 | for(int i=0;i>arr[i];
28 |
29 |
30 | for(int i=0;i
2 | using namespace std;
3 | // Question:
4 | // Sort the elements of the array in non-decreasing order using insertion sort.
5 |
6 | // This function is called by reference.
7 | void insertionSort(vector& nums)
8 | {
9 | int i, key, j, n=nums.size();
10 | for (i = 1; i < n; i++)
11 | {
12 | // To sort the array we will first pick the ith element as key.
13 | // And find the proper position for it.
14 | // The position before that all elements in the array are smaller than key.
15 | // We will shift all the elements from that position to right of it.
16 | key = nums[i];
17 | j = i - 1;
18 |
19 | // This while loop will shift the elements in right.
20 | while (j >= 0 && nums[j] > key)
21 | {
22 | // j+1th element takes the value of jth element
23 | nums[j + 1] = nums[j];
24 | j = j - 1;
25 | }
26 |
27 | // The loop will stop once we find the correct position.
28 | // And j+1 is the correct position for that particular element.
29 | nums[j + 1] = key;
30 | }
31 | }
32 |
33 | int main()
34 | {
35 | int n;
36 | cin>>n;
37 |
38 | vector nums(n);
39 | for(int i=0; i>nums[i];
41 | }
42 |
43 | insertionSort(nums);
44 | for(int i=0; i
13 | using namespace std;
14 |
15 | int main()
16 | {
17 |
18 | //number of elements
19 | int n;
20 | cin>>n;
21 |
22 | int arr[n];
23 | for(int i=0;i>arr[i];
25 |
26 | for(int i=0;i
18 | # include
19 | # include
20 | using namespace std;
21 |
22 | int main() {
23 |
24 | int n,x,i,j;
25 | cin>>n;
26 | int array[n];
27 |
28 | // vector even stores even index elements of array.
29 | // vector odd stores odd index elements of array.
30 |
31 | vector