└── README.md


/README.md:
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  1 | ## Problem 1:
  2 | Given 3 tables, students(id, name) , courses(id, name) , grades(id, course_id, student_id, grade), 
  3 | 
  4 | find the top 100 students based on their 
  5 | 
  6 | average grades sorted descendingly by the average grade and in case multiple students have the same average grade, 
  7 | sort them lexicographically in ascending order by their names. 
  8 | 
  9 | Your query should output a table with the following columns (name, average_grade).
 10 | 
 11 | #### Solution:
 12 | ```SQL
 13 | SELECT DISTINCT students.name AS name, avg(grade) AS average_grade FROM grades
 14 | RIGHT JOIN students ON students.id = grades.student_id
 15 | GROUP BY students.name
 16 | ORDER BY avg(grade) DESC, students.name ASC
 17 | LIMIT 100
 18 | ```
 19 | 
 20 | ___
 21 | ## Problem 2:
 22 | 
 23 | For the same tables in **Problem 1**, for each student, get all the courses that he/she is enrolled in along with the grade he/she scored for each course.
 24 | Order the result by the student name in ascending order and if there is a tie, break it with the course name in ascending order and
 25 | if there is a tie break it with the grade in ascending order. The final result should have 3 columns with names (name, course, grade).
 26 | 
 27 | #### Solution:
 28 | 
 29 | ```SQL
 30 | SELECT DISTINCT students.name AS name,courses.name AS course, grades.grade AS grade 
 31 | FROM ((grades
 32 | INNER JOIN students ON students.id = grades.student_id
 33 | INNER JOIN courses ON courses.id = grades.course_id))
 34 | ORDER BY students.name ASC, courses.name ASC, grades.grade ASC
 35 | ```
 36 | 
 37 | ___
 38 | 
 39 | ## Problem 3:
 40 | For the same tables in **Problem 1**, get the name of the most popular course (the one where the most students are enrolled) and if there is a tie, get the course that's lexicographically the smallest.
 41 | #### Solution:
 42 | ```SQL
 43 | SELECT c.name
 44 | FROM courses c
 45 | INNER JOIN grades g ON g.course_id = c.id
 46 | GROUP BY c.id, c.name
 47 | ORDER BY COUNT(*) DESC, c.name
 48 | LIMIT 1;
 49 | ```
 50 | ___
 51 | ## Problem 4:
 52 | Given a table called "bugs" with the following columns (id, token, title, category, device, reported_at, created_at, updated_at). 
 53 | Select all distinct bug categories. 
 54 | #### Solution:
 55 | ```SQL
 56 | SELECT DISTINCT category FROM bugs
 57 | ```
 58 | ___
 59 | 
 60 | ## Problem 5: 
 61 | For the same table in **Problem 4**, find how many bugs were created on "2019-03-01" or later. Your query should produce a table with one column called "count". 
 62 | This problem is graded partially, 10% on correctness (your query gets the correct count) and 90% on performance (your query makes use of available indexes).
 63 | 
 64 | ***Scored 10% only***
 65 | "Not Optimal solution"
 66 | 
 67 | ```SQL
 68 | SHOW INDEXES FROM bugs;
 69 | ```
 70 | 
 71 | ```JSON
 72 | "records": [
 73 | 		{
 74 | 			"Table": "bugs",
 75 | 			"Non_unique": 0,
 76 | 			"Key_name": "PRIMARY",
 77 | 			"Seq_in_index": 1,
 78 | 			"Column_name": "id",
 79 | 			"Collation": "A",
 80 | 			"Cardinality": 9791826,
 81 | 			"Sub_part": null,
 82 | 			"Packed": null,
 83 | 			"Null": "",
 84 | 			"Index_type": "BTREE",
 85 | 			"Comment": "",
 86 | 			"Index_comment": ""
 87 | 		},
 88 | 		{
 89 | 			"Table": "bugs",
 90 | 			"Non_unique": 1,
 91 | 			"Key_name": "index_bugs_on_category_and_token_and_reported_at",
 92 | 			"Seq_in_index": 1,
 93 | 			"Column_name": "category",
 94 | 			"Collation": "A",
 95 | 			"Cardinality": 1,
 96 | 			"Sub_part": null,
 97 | 			"Packed": null,
 98 | 			"Null": "YES",
 99 | 			"Index_type": "BTREE",
100 | 			"Comment": "",
101 | 			"Index_comment": ""
102 | 		},
103 | 		{
104 | 			"Table": "bugs",
105 | 			"Non_unique": 1,
106 | 			"Key_name": "index_bugs_on_category_and_token_and_reported_at",
107 | 			"Seq_in_index": 2,
108 | 			"Column_name": "token",
109 | 			"Collation": "A",
110 | 			"Cardinality": 29946,
111 | 			"Sub_part": null,
112 | 			"Packed": null,
113 | 			"Null": "YES",
114 | 			"Index_type": "BTREE",
115 | 			"Comment": "",
116 | 			"Index_comment": ""
117 | 		},
118 | 		{
119 | 			"Table": "bugs",
120 | 			"Non_unique": 1,
121 | 			"Key_name": "index_bugs_on_category_and_token_and_reported_at",
122 | 			"Seq_in_index": 3,
123 | 			"Column_name": "reported_at",
124 | 			"Collation": "A",
125 | 			"Cardinality": 6085027,
126 | 			"Sub_part": null,
127 | 			"Packed": null,
128 | 			"Null": "YES",
129 | 			"Index_type": "BTREE",
130 | 			"Comment": "",
131 | 			"Index_comment": ""
132 | 		}
133 | 	]
134 | }
135 | ```
136 | ```SQL
137 | SELECT COUNT(x.created_at) AS count 
138 | FROM bugs AS y 
139 | USE INDEX (PRIMARY,index_bugs_on_category_and_token_and_reported_at) 
140 | INNER JOIN bugs AS x ON x.id=y.id 
141 | WHERE x.created_at >= '2019-03-01
142 | ```
143 | 
144 | ___
145 | 
146 | ## Problem 6:
147 | 
148 | For the same table in **Problem 4**, find the title of the bug with token = "token660" and reported_at on "2020-08-30". 
149 | This problem is graded partially, 10% on correctness (your query gets the correct count) and 90% on performance (your query makes use of available indexes). 
150 | 
151 | ***Scored 10% only*** "Not Optimal Solution"
152 | ```SQL
153 | SELECT x.title FROM bugs
154 | AS y USE INDEX (index_bugs_on_category_and_token_and_reported_at)
155 | JOIN bugs AS x ON x.id=y.id  
156 | WHERE y.token ='token660' AND y.reported_at = '2020-08-30' 
157 | ```
158 | 


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