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Author: ggeerraarrdd

README.md

@@ -144,6 +144,7 @@ To make the daily tasks of creating a new file and updating the index easier, th
 | 113   | [Queries Quality and Percentage](https://leetcode.com/problems/queries-quality-and-percentage/description/)                                                   | [Solution](solutions/113_queries_quality_and_percentage.md)                           | LeetCode    | Easy         |   |
 | 114   | [Monthly Transactions I](https://leetcode.com/problems/monthly-transactions-i/description/)                                                                   | [Solution](solutions/114_monthly_transactions_i.md)                                   | LeetCode    | Medium       |   |
 | 115   | [Immediate Food Delivery II](https://leetcode.com/problems/immediate-food-delivery-ii/description/)                                                           | [Solution](solutions/115_immediate_food_delivery_ii.md)                               | LeetCode    | Medium       |   |
+| 116   | [Game Play Analysis IV](https://leetcode.com/problems/game-play-analysis-iv/description/)                                                                     | [Solution](solutions/116_game_play_analysis_iv.md)                                    | LeetCode    | Medium       |   |
 
 ## Author(s)
 

solutions/116_game_play_analysis_iv.md

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+# SQL Everyday \#116
+
+## Game Play Analysis IV
+
+Site: LeetCode\
+Difficulty per Site: Medium
+
+## Problem
+
+Write a solution to report the *fraction* of players that logged in again on the day after the day they first logged in, *rounded to 2 decimal places*. In other words, you need to count the number of players that logged in for at least two consecutive days starting from their first login date, then divide that number by the total number of players. [[Full Description](https://leetcode.com/problems/game-play-analysis-iv/description/)]
+
+## Submitted Solution
+
+```sql
+-- Submitted Solution
+WITH cte AS
+    (
+    SELECT
+        player_id
+        ,DATE(MIN(event_date) + INTERVAL 1 DAY) AS second_day
+    FROM Activity
+    GROUP BY player_id
+    )
+SELECT
+    ROUND(COUNT(*) / (SELECT COUNT(DISTINCT player_id) FROM Activity), 2) AS fraction
+FROM Activity a
+JOIN cte ON a.player_id = cte.player_id AND a.event_date = cte.second_day
+;
+```
+
+## Site Solution
+
+```sql
+-- LeetCode Solution 
+-- Approach 1: Subqueries and multi-value use of the IN comparison operator
+SELECT
+  ROUND(
+    COUNT(A1.player_id)
+    / (SELECT COUNT(DISTINCT A3.player_id) FROM Activity A3)
+  , 2) AS fraction
+FROM
+  Activity A1
+WHERE
+  (A1.player_id, DATE_SUB(A1.event_date, INTERVAL 1 DAY)) IN (
+    SELECT
+      A2.player_id,
+      MIN(A2.event_date)
+    FROM
+      Activity A2
+    GROUP BY
+      A2.player_id
+  );
+  
+-- Approach 2: CTEs and INNER JOIN
+WITH first_logins AS (
+  SELECT
+    A.player_id,
+    MIN(A.event_date) AS first_login
+  FROM
+    Activity A
+  GROUP BY
+    A.player_id
+), consec_logins AS (
+  SELECT
+    COUNT(A.player_id) AS num_logins
+  FROM
+    first_logins F
+    INNER JOIN Activity A ON F.player_id = A.player_id
+    AND F.first_login = DATE_SUB(A.event_date, INTERVAL 1 DAY)
+)
+SELECT
+  ROUND(
+    (SELECT C.num_logins FROM consec_logins C)
+    / (SELECT COUNT(F.player_id) FROM first_logins F)
+  , 2) AS fraction;
+```
+
+## Notes
+
+TODO
+
+Go to [Table of Contents](/README.md#contents)\
+Go to [Overview](/README.md)