SQL2

DepartmentTop3Salarries.sql

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+# Write your MySQL query statement below
+/*
+SELECT d.name AS 'Department',e1.name as 'Employee', e1.salary 
+FROM Employee e1
+LEFT JOIN Department d 
+ON e1.departmentId = d.id
+WHERE 3 > (SELECT COUNT(DISTINCT e2.salary) FROM Employee e2 WHERE e2.salary > e1.salary AND e1.departmentId = e2.departmentId ) 
+*/
+
+
+
+WITH CTE AS 
+(SELECT e1.name , e1.departmentId,e1.salary, DENSE_RANK() OVER(PARTITION BY e1.departmentId ORDER BY e1.salary DESC) as 'rank' FROM Employee e1 )
+SELECT  d.name AS Department , C.name as 'Employee', C.salary AS 'salary' FROM CTE C
+JOIN Department d ON C.departmentID = d.id
+WHERE C.rank < 4;

RankScores.sql

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+# Write your MySQL query statement below
+ /* My Approach
+WITH CTE_TABLE AS (
+    SELECT score, ROW_NUMBER() OVER(ORDER BY score DESC) as RNK  FROM Score
+)
+
+SELECT s1.score, c1.Rnk 
+FROM Score s1
+Join CTE_TABLE c1 ON s1.score = c1.score; */
+
+
+
+SELECT score, DENSE_RANK() OVER(ORDER BY score DESC) AS 'rank'
+FROM Scores

SeatSwap.sql

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+
+
+
+SELECT(
+    CASE 
+        WHEN MOD(ID,2) !=0 AND ID != COUNTER THEN ID+1
+        WHEN MOD(ID,2) !=0 AND ID = COUNTER THEN ID
+        ELSE ID -1
+    END
+) AS 'id', student FROM Seat , ( SELECT COUNT(*) as COUNTER FROM Seat) AS seat_count  ORDER BY id;

TreeNode.sql

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+
+
+SELECT id,(
+         CASE 
+    WHEN p_id IS NULL THEN 'Root'
+    WHEN id IN (SELECT p_id FROM Tree)  THEN 'Inner'
+    ELSE  'Leaf'
+ END) as Type FROM Tree;