├── README.md
├── Interesting Alphabets.py
└── Total Salary.py


/README.md:
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1 | # 6-Month-Code-Marathon---1


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/Interesting Alphabets.py:
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 1 | """Problem statement
 2 | As a part of its competition, the school will conduct a codeathon, Lock the Code, where it has been given a value, and the participants have to decode it.
 3 | 
 4 | The participants are given a value denoting the number of rows in the matrix; they need to print the pattern.
 5 | 
 6 | Example : 
 7 | 
 8 | For N=5, Pattern:
 9 | E
10 | DE
11 | CDE
12 | BCDE
13 | ABCDE 
14 | Among the participants, Ninja is new to programming and doesn’t have much experience; he asks you to solve the problem. Can you help solve this problem?
15 | 
16 | Detailed explanation ( Input/output format, Notes, Images )
17 | Constraints:
18 | 1 <= T <= 50
19 | 1 <= N <= 26
20 | 
21 | Time Limit: 1 sec
22 | Sample Input 1:
23 | 2
24 | 5
25 | 4
26 | Sample Output 1:
27 | E
28 | DE
29 | CDE
30 | BCDE
31 | ABCDE
32 | 
33 | D
34 | CD
35 | BCD
36 | ABCD
37 | Explanation for Sample Input 1:
38 | In the first test case, value of ‘N’ is 5, so print the ‘N’ rows from 1 to ‘N’ where in each row start from (N - i - 1)the character which goes on till ‘Nth character. Hence the answer is [‘E’,’DE’,’CDE,’ BCDE’,’ABCDE’].
39 | 
40 | In the second test case, the value of ‘N’ is 4, so print the ‘N’ rows from 1 to ‘N’ where each row starts from (N - i - 1)the character, which goes on till ‘Nth character. Hence the answer is [‘D’,’CD’,BCD’,’ABCD’].
41 | Sample Input 2:
42 | 2
43 | 3
44 | 2
45 | Sample Output 2:
46 | C
47 | BC
48 | ABC
49 | 
50 | B
51 | AB"""
52 | 
53 | def interestingPattern(n):
54 |     result = []
55 |     for i in range(n):
56 |         start_char = chr(ord('A') + n - i - 1)
57 |         row = ' '.join([chr(j) for j in range(ord(start_char), ord('A') + n)])
58 |         result.append(row)
59 |     return result
60 | 


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/Total Salary.py:
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 1 | """Problem statement
 2 | Ninja just got an offer letter from a reputable company. The company sent him an offer letter along with the salary bifurcation.
 3 | 
 4 | In that bifurcation,Total Salary was not mentioned but instead a ‘basicSalary’ and an upper case character representing grade was mentioned, depending on which the Total Salary is calculated.
 5 | 
 6 | Help Ninja in calculating his total salary, where total salary is defined as:
 7 | 
 8 | ‘totalSalary’ = ‘basic’ + ‘hra’ + ‘da’ + ‘allowance’ - ‘pf’
 9 | The above terms are as follows:
10 | 
11 | ‘hra’ = 20% of ‘basic’
12 | ‘da’ = 50% of ‘basic’
13 | ‘allowance’ = 1700 if grade = ‘A’
14 | ‘allowance’ = 1500 if grade = ‘B’
15 | ‘allowance’ = 1300 if grade = ‘C' or any other character
16 | ‘pf’ = 11% of ‘basic’.
17 | Note :
18 | 
19 | Round off the ‘totalSalary’ and then print the integral part only.
20 | 
21 | 'x.5' type values will always be round up, for example, 1.5, 2.5 will be round off to 2, 3 respectively.
22 | Detailed explanation ( Input/output format, Notes, Images )
23 | Constraints :
24 | 0 <= 'basicSalary' <= 7 * (10 ^ 5)
25 | 
26 | Time Limit: 1 sec.
27 | Sample Input 1 :
28 | 2
29 | 10000 A
30 | 4567 B
31 | Sample Output 1 :
32 | 17600
33 | 8762
34 | Explanation for Sample Input 1:
35 | Test Case 1:
36 | We have been given the basic salary as Rs. 10000. We need to calculate the hra, da and pf. 
37 | Now when we calculate each of the, it turns out to be:
38 | hra =  20% of Rs. 10000 = Rs. 2000
39 | da = 50% od Rs. 10000 = Rs. 5000
40 | pf = 11% of Rs. 10000 = Rs. 1100
41 | 
42 | Since, the grade is 'A', we take an allowance of Rs. 1700.
43 | On substituting these values to the formula of totalSalary, we get Rs. 17600 and now rounding it off will result in Rs. 17600 and hence the Answer.
44 | 
45 | Test Case 2:
46 | We have been given the basic salary as Rs. 4567. We need to calculate the hra, da and pf. 
47 | Now when we calculate each of the, it turns out to be:
48 | hra =  20% of Rs. 4567 = Rs. 913.4
49 | da = 50% od Rs. 4567 = Rs. 2283.5
50 | pf = 11% of Rs. 4567 = Rs. 502.37
51 | 
52 | Since, the grade is 'B', we take an allowance of Rs. 1500.
53 | On substituting these values to the formula of totalSalary, we get Rs. 8761.53 and now rounding it off will result in Rs. 8762 and hence the Answer.
54 | Sample Input 2 :
55 | 2
56 | 1500 B
57 | 5000 C
58 | Sample Output 2 :
59 | 3885
60 | 9250 """
61 | 
62 | 
63 | 
64 | 
65 | from math import floor
66 | 
67 | def totalSalary(basic, grade):
68 |     hra = 0.2 * basic
69 |     da = 0.5 * basic
70 |     pf = 0.11 * basic
71 | 
72 |     if grade == 'A':
73 |         allowance = 1700
74 |     elif grade == 'B':
75 |         allowance = 1500
76 |     else:
77 |         allowance = 1300
78 | 
79 |     total = basic + hra + da + allowance - pf
80 |     return floor(total + 0.5)  # Round .5 up
81 | 


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