SQL5 problem solutions

AverageSalaryDepartmentvsCompany.sql

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+-- Solution for Leetcode problem AverageSalaryDepartmentvsCompany
+WITH dept AS (
+    SELECT  DATE_FORMAT(s.pay_date, '%Y-%m') as pay_month,
+            e.department_id,
+            AVG(s.amount) AS d_avg
+    FROM salary s
+    JOIN employee e
+    ON s.employee_id = e.employee_id
+    GROUP BY 1,2
+),
+comp AS(
+    SELECT  DATE_FORMAT(pay_date, '%Y-%m') as pay_month,
+            AVG(amount) AS c_avg
+    FROM salary
+    GROUP BY 1
+)
+SELECT  d.pay_month,
+        d.department_id,
+        CASE
+            WHEN d.d_avg > c.c_avg THEN 'higher'
+            WHEN d.d_avg < c.c_avg THEN 'lower'
+            WHEN d.d_avg = c.c_avg THEN 'same'
+        END AS comparison
+FROM dept d
+JOIN comp c
+ON d.pay_month = c.pay_month;

Report Contiguos Dates.sql

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+WITH cte AS (
+    SELECT  fail_date AS date,
+            'failed' AS period_state,
+            RANK() OVER(ORDER BY fail_date) rnk
+    FROM failed
+    WHERE fail_date BETWEEN '2019-01-01' AND '2019-12-31'
+    UNION ALL
+    SELECT  success_date AS date,
+            'succeeded' AS period_state,
+            RANK() OVER(ORDER BY success_date) rnk
+    FROM succeeded
+    WHERE success_date BETWEEN '2019-01-01' AND '2019-12-31'
+)
+SELECT  period_state,
+        MIN(date) AS start_date,
+        MAX(date) AS end_date
+FROM (
+    SELECT  *,
+            RANK() OVER(ORDER BY date) main_rnk,
+            RANK() OVER(ORDER BY date) - rnk AS group_col -- this is done to identify the difference between the ranks.
+            -- which helps to group. All records with similar difference value is considered in same group.
+    FROM cte
+    ORDER BY date
+) a
+GROUP BY period_state, group_col;

Student Report By Geography.sql

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+-- Method 1
+WITH america AS (
+    SELECT  name AS america,
+            ROW_NUMBER() OVER(ORDER BY name) as rwn
+    FROM student
+    WHERE continent = 'America'
+),
+asia AS (
+    SELECT  name AS asia,
+            ROW_NUMBER() OVER(ORDER BY name) as rwn
+    FROM student
+    WHERE continent = 'Asia'
+),
+europe AS (
+    SELECT  name AS europe,
+            ROW_NUMBER() OVER(ORDER BY name) as rwn
+    FROM student
+    WHERE continent = 'Europe'
+)
+SELECT  am.america,
+        a.asia,
+        e.europe
+FROM america am
+LEFT JOIN asia a-- left join because the questions says that America has more number of students
+ON am.rwn = a.rwn
+LEFT JOIN europe e
+ON am.rwn = e.rwn;
+
+
+-- Method 2 - POSTGRES QUERY - In case it is not clear which country has more number of students (Works in Postgres not MySQL)
+WITH america AS (
+    SELECT  name AS america,
+            ROW_NUMBER() OVER(ORDER BY name) as rwn
+    FROM student
+    WHERE continent = 'America'
+),
+asia AS (
+    SELECT  name AS asia,
+            ROW_NUMBER() OVER(ORDER BY name) as rwn
+    FROM student
+    WHERE continent = 'Asia'
+),
+europe AS (
+    SELECT  name AS europe,
+            ROW_NUMBER() OVER(ORDER BY name) as rwn
+    FROM student
+    WHERE continent = 'Europe'
+)
+SELECT  am.america,
+        a.asia,
+        e.europe
+FROM america am
+FULL JOIN asia a
+ON am.rwn = a.rwn
+FULL JOIN europe e
+ON am.rwn = e.rwn;
+
+-- Method 3 - Using the session variables isntead of window funtion ROW_NUMBER
+SELECT  am.america,
+        a.asia,
+        e.europe
+FROM (
+    (SELECT @america := 0, @asia := 0, @europe := 0) sess_var,
+    (SELECT  @america := @america + 1 AS rwn,
+            name AS america
+    FROM student
+    WHERE continent = 'America'
+    ORDER BY name) am
+    LEFT JOIN 
+    (SELECT  @asia := @asia + 1 AS rwn,
+            name AS asia
+    FROM student
+    WHERE continent = 'Asia'
+    ORDER BY name) a
+    ON am.rwn = a.rwn
+    LEFT JOIN 
+    (SELECT  @europe := @europe + 1 AS rwn,
+            name AS europe
+    FROM student
+    WHERE continent = 'Europe'
+    ORDER BY name) e
+    ON am.rwn = e.rwn
+);

game_analysis_1

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+-- Method 1
+SELECT  player_id,
+        MIN(event_date) first_login
+FROM activity
+GROUP BY 1;
+
+-- Method 2
+SELECT  DISTINCT player_id,
+        FIRST_VALUE(event_date) OVER(PARTITION BY player_id ORDER BY event_date RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) first_login
+FROM activity;
+
+-- Method 3
+SELECT  player_id,
+        event_date AS first_login
+FROM (
+    SELECT  player_id,
+            event_date,
+            DENSE_RANK() OVER(PARTITION BY player_id ORDER BY event_date) rnk
+    FROM activity
+)a
+WHERE rnk = 1