MySQL Solution for Game Play Analysis IV Problem
Intuition\r\n Describe your first thoughts on how to solve this problem. \r\nThe solution utilizes two common table expressions (CTEs) to break down the problem
# Intuition\r\n<!-- Describe your first thoughts on how to solve this problem. -->\r\nThe solution utilizes two common table expressions (CTEs) to break down the problem into smaller steps. The first CTE, first_logins, finds the first login date for each player. The second CTE, consec_logins, counts the number of consecutive logins for each player. Finally, the main query combines the results from these CTEs to calculate the desired fraction.\r\n\r\n# Approach\r\n<!-- Describe your approach to solving the problem. -->\r\n1. The first_logins CTE selects the player_id and the minimum event_date for each player using the GROUP BY clause and the MIN function.\r\n1. The consec_logins CTE joins the first_logins CTE with the Activity table, matching the player_id and filtering the records where the event_date of the second login is exactly one day before the first login. It then counts the number of matching records for each player using the COUNT function.\r\n1. The main query retrieves the num_logins from the consec_logins CTE and divides it by the count of distinct player_ids from the first_logins CTE. The result is rounded to two decimal places using the ROUND function.\r\n\r\n# Complexity\r\n- Time complexity:\r\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\r\nThe time complexity of this solution depends on the size of the Activity table and the efficiency of the underlying database\'s query optimizer. The first CTE, first_logins, performs a simple grouping operation, which typically has a time complexity of O(n log n) or better, where n is the number of rows in the Activity table. The second CTE, consec_logins, involves joining the Activity table, which can have a time complexity of O(n^2) in the worst case. However, the performance can be improved with appropriate indexing on the relevant columns. The overall time complexity of the solution is determined by the more time-consuming of these operations.\r\n\r\n- Space complexity:\r\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\r\nThe space complexity of this solution depends on the number of distinct player_ids in the Activity table. The first_logins CTE stores the player_id and the corresponding first_login date for each player, which requires space proportional to the number of distinct players. Similarly, the consec_logins CTE stores the number of consecutive logins for each player, which also requires space proportional to the number of distinct players. Therefore, the space complexity is O(m), where m is the number of distinct players.\r\n\r\n# Code\r\n```\r\n# Write your MySQL query statement below\r\n# SELECT ROUND(COUNT(DISTINCT t1.player_id) / COUNT(DISTINCT t2.player_id), 2) AS fraction\r\n# FROM Activity AS t1\r\n# JOIN Activity AS t2 ON t1.player_id = t2.player_id AND t1.event_date = DATE_SUB(t2.event_date, INTERVAL 1 DAY)\r\n# GROUP BY t1.player_id\r\n# HAVING COUNT(DISTINCT t1.event_date) > 1;\r\n# SELECT ROUND(COUNT(DISTINCT t1.player_id) / (SELECT COUNT(DISTINCT player_id) FROM Activity), 2) AS fraction\r\n# FROM Activity AS t1\r\n# WHERE EXISTS (\r\n#     SELECT *\r\n#     FROM Activity AS t2\r\n#     WHERE t1.player_id = t2.player_id\r\n#     AND t2.event_date = DATE_ADD(t1.event_date, INTERVAL 1 DAY)\r\n# );\r\n\r\nWITH first_logins AS (\r\n  SELECT\r\n    A.player_id,\r\n    MIN(A.event_date) AS first_login\r\n  FROM\r\n    Activity A\r\n  GROUP BY\r\n    A.player_id\r\n), consec_logins AS (\r\n  SELECT\r\n    COUNT(A.player_id) AS num_logins\r\n  FROM\r\n    first_logins F\r\n    INNER JOIN Activity A ON F.player_id = A.player_id\r\n    AND F.first_login = DATE_SUB(A.event_date, INTERVAL 1 DAY)\r\n)\r\nSELECT\r\n  ROUND(\r\n    (SELECT C.num_logins FROM consec_logins C)\r\n    / (SELECT COUNT(F.player_id) FROM first_logins F)\r\n  , 2) AS fraction;\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n```
['MySQL']