completed sql2

PRoblem4.sql

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+WITH cte AS (
+    SELECT 
+        d.name AS Department, 
+        e.name AS Employee, 
+        e.salary AS Salary,
+        DENSE_RANK() OVER (
+            PARTITION BY e.departmentId 
+            ORDER BY e.salary DESC
+        ) AS salary_rank
+    FROM Employee e 
+    LEFT JOIN Department d 
+        ON e.departmentId = d.id
+)
+SELECT 
+    Department, 
+    Employee, 
+    Salary
+FROM cte 
+WHERE salary_rank <= 3
+ORDER BY Department, Salary DESC;
+

Problem1.sql

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+# Write your MySQL query statement below
+-- SELECT s1.score, (select COUNT(DISTINCT s2.score)
+-- FROM scores s2 WHERE s2.score>=s1.score )AS 'rank'
+-- FROM scores s1 ORDER BY s1.score DESC;
+
+SELECT s.score, COUNT(DISTINCT T.score) AS 'rank'
+FROM scores s JOIN scores T ON s.score<=T.score
+GROUP BY s.id ORDER BY S.SCORE DESC;

Problem2.sql

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+# Write your MySQL query statement below
+-- SELECT  count(id) from FROM seats;
+-- SELECT MOD(id,2) AS 'mod' 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, (SELECT COUNT(*) as 'cnts' FROM Seat) AS SEAT_COUNTS  ORDER BY ID;
+select s1.id, COALESCE (s2.student, s1.student) AS'student' FROM seat s1 LEFT JOIN seat s2 ON (s1.id +1)^1-1 = s2.id ORDER BY s1.id;

Problem3.sql

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+# Write your MySQL query statement below
+
+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) AND p_id is NOT NULL THEN 'Leaf'
+    ELSE 'Inner'
+    END
+) AS 'type' FROM tree;