From ba257262e926274c0b58f100e10e7655af8faa2e Mon Sep 17 00:00:00 2001
From: Kaustubhk28 <99148707+Kaustubhk28@users.noreply.github.com>
Date: Mon, 23 Dec 2024 15:33:21 -0500
Subject: [PATCH] Done Sql2

---
 Sql2.sql | 121 +++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 121 insertions(+)
 create mode 100644 Sql2.sql

diff --git a/Sql2.sql b/Sql2.sql
new file mode 100644
index 0000000..19ebc45
--- /dev/null
+++ b/Sql2.sql
@@ -0,0 +1,121 @@
+# Problem 1 : Rank Scores (https://leetcode.com/problems/rank-scores/ )
+
+# 1st Solution
+SELECT score, DENSE_RANK() OVER(ORDER BY score DESC) AS 'rank' 
+FROM Scores
+
+# 2nd Solution
+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
+
+# 3rd Solution
+SELECT s.score, COUNT(DISTINCT t.score) AS 'rank'
+FROM Scores s
+INNER JOIN Scores t
+ON s.score <= t.score
+GROUP BY s.score, s.id
+ORDER BY s.score DESC
+
+# Problem 2 : Exchange Seats (https://leetcode.com/problems/exchange-seats/ )
+
+# 1st solution
+SELECT 
+    CASE 
+        WHEN id % 2 = 1 AND id != cnt THEN id+1
+        WHEN id % 2 = 1 AND id = cnt THEN id
+        ELSE id-1
+    END AS 'id', student 
+FROM Seat, (SELECT COUNT(id) AS 'cnt' FROM Seat) AS seat_count
+ORDER BY id;
+
+# 2nd solution
+SELECT 
+    CASE 
+        WHEN id % 2 = 1 AND id = (SELECT MAX(id) FROM seat) THEN id
+        WHEN id % 2 = 1 THEN id+1
+        ELSE id-1
+    END AS 'id', student 
+FROM Seat
+ORDER BY id;
+
+# 3rd solution
+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;
+
+# Problem 3 : Tree Node (https://leetcode.com/problems/tree-node/ )
+
+# 1st Solution
+SELECT id, 
+    CASE
+        WHEN p_id IS NULL THEN 'Root'
+        WHEN id NOT IN (SELECT p_id FROM Tree WHERE p_id IS NOT NULL) THEN 'Leaf'
+        ELSE 'Inner'
+    END as 'type'
+FROM Tree    
+
+# 2nd Solution
+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    
+
+# 3rd Solution
+    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 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 p_id
+        FROM Tree
+        WHERe p_id IS NOT NULL
+    ) AND p_id IS NOT NULL
+
+# 4th Solution
+SELECT id, 
+    IF (p_id IS NULL, 'Root' , IF(id IN (SELECT p_id FROM Tree), 'Inner', 'Leaf')) AS type
+FROM Tree
+
+# Problem 4 : Deparment Top 3 Salaries (https://leetcode.com/problems/department-top-three-salaries/ )
+
+# 1st Solution
+WITH temp AS
+(
+    SELECT d.name AS Department, e.name AS Employee, e.salary AS Salary, 
+        DENSE_RANK() OVER(PARTITION BY d.id ORDER BY e.salary DESC) AS ranks
+    FROM Employee e
+    JOIN Department d
+    ON e.departmentId = d.id
+)
+SELECT Department, Employee, Salary
+FROM temp
+WHERE ranks IN (1,2,3)
+
+# 2nd Solution
+SELECT d.name AS Department, e.name AS Employee, e.salary AS Salary 
+FROM Employee e
+JOIN Department d
+ON e.departmentId = d.id
+WHERE 3 > (SELECT COUNT(DISTINCT e2.salary) 
+            FROM Employee e2 
+            WHERE e.salary < e2.salary AND e.departmentId = e2.departmentId)