Easy MySQL Solution
PLEASE UPVOTE\n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nWE HAVE TO CHOOSE MINIMUM DATE + DATE AFTER.\n\n# Approach\n Describ
# PLEASE UPVOTE\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWE HAVE TO CHOOSE MINIMUM DATE + DATE AFTER.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nMAKE A TABLE WHICH CONTAINS MINIMUM DATES, as A SUBQUERY\nLIKE THIS\n```\nSelect player_id,min(event_date) as ed1 from Activity \ngroup by player_id\n```\n\nand where statement as\n```\nWHERE a.event_date=s.ed1+INTERVAL 1 DAY\n```\n\n# Complexity\n- Time complexity:O(N^2)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```mysql []\n# Write your MySQL query statement below\n\nWith subq1 as(\n    Select player_id,min(event_date) as ed1 from Activity \n    group by player_id\n)\n\nSELECT Round(Count(DISTINCT a.player_id)/(select count(distinct player_id) from Activity),2) as fraction\nFROM Activity a left join subq1 s\non a.player_id=s.player_id\nWHERE a.event_date=s.ed1+INTERVAL 1 DAY;\n```
['Database', 'MySQL']