completed sql2 problems

01-RankScores.sql

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+select score, dense_rank()
+over (order by score desc) as 'rank'
+from scores 

02-ExchangeSeat.sql

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+-- Problem 2 : Exchange Seats	(https://leetcode.com/problems/exchange-seats/ )
+
+-- solution from the class
+select (
+    case 
+        when mod(id,2) !=0 and id=cnts then id 
+        when mod(id,2) !=0 and id!=cnts then id+1
+        else id-1
+    end
+) as 'id', student from seat, 
+(select count(*) as 'cnts' from seat) as seat_counts order by id
+
+-- I found the above counter intuitive, so I tried this:
+with seat_counts as (select count(*) as cnts from seat)
+select (
+    case
+        when mod(id,2) != 0 and id!=cnts then id+1
+        when mod(id,2) != 0 and id=cnts then id
+        else id-1 
+    end
+    ) as id, student 
+from seat, seat_counts
+order by id
+
+-- i also thought of the following but realised it will be very slow 
+-- because the it will run the subquery select max(id) multiple times
+
+SELECT (
+    CASE 
+        WHEN MOD(id, 2) != 0 AND id = (SELECT MAX(id) FROM seat) THEN id      
+        WHEN MOD(id, 2) != 0 AND id != (SELECT MAX(id) FROM seat) THEN id + 1 
+        ELSE id - 1                                                           
+    END
+) AS 'id', student
+FROM seat
+ORDER BY id;
+
+-- solution using XOR 
+
+select s1.id, ifnull(s2.student, s1.student) as student
+from seat s1 left join seat s2 on (s1.id+1)^1 - 1 = s2.id
+
+
+/*
+
+- idea: expression is (x+1)^1 - 1 which return one less when even and one more when odd (check notes)
+check id, say 1 -> expression returns 2 -> go to copy, search name belonging to this value and put it with 1
+create two copies of the table and make join 
+- always be careful when updating the same table that the previous value is not overwritten, therfore create copy of the table
+
+-- coalesce can be used instead of ifnull, it can handle multiple null cases one by one
+-- i think order by is not needed here because of how id is provided (increment continuously)
+
+select (id+1)^1 - 1 as id, student
+from seat 
+order by id
+
+only problem is the case with odd number of rows 
+*/

03-TreeNode.sql

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+-- Problem 3 : Tree Node (https://leetcode.com/problems/tree-node/)
+
+/*
+solved before class, thought process: 
+- when pid is null then root
+- when both in id and pid then inner 
+- if only in id then leaf
+*/
+select id, (
+    case 
+        when p_id is null then 'Root'
+        when id not in (select distinct p_id from tree where p_id is not null)  then 'Leaf'
+        else 'Inner'
+    end
+) as 'type'
+from tree
+
+-- from class 
+select id, 'Root' as type from tree where p_id is null
+union 
+select id, 'Leaf' as type from tree where id not in
+    (select distinct p_id from tree where p_id is not null) and p_id is not null 
+union
+select id, 'Inner' as type from tree where id in
+    (select distinct p_id from tree) and p_id is not null 
+
+-- case was also covered in class and same solution
+
+-- with nested if 
+select id, 
+if ( isnull(p_id), 'Root',
+    if(id in (select distinct p_id from tree), 'Inner', 'Leaf'
+    )
+) as type from tree
+
+/*
+- if (expression, true, false)
+- if (expression1, true, if (expression2, true, false) )
+
+- short-circuiting: in both case and nested if the condition 
+is satifies in 1st case it will bot be checked in the next one
+but in union it is not the case so we have to write all the conditions again (p_id is not null)
+
+- union will return single entry however union all will return all the values with duplicated value 
+as well
+*/

04-DepartmentTop3Salaries.sql

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+-- Problem 4 : Deparment Top 3 Salaries	(https://leetcode.com/problems/department-top-three-salaries/)
+
+with cte as (select d.name as Department, e.name as Employee, salary, 
+dense_rank() over (partition by d.name order by salary desc) as rnk
+from employee e 
+left join department d 
+on e.departmentId = d.id)
+select Department, Employee, salary 
+from cte where rnk <=3