SOLUTION WITH SUM() OVER() WINDOW FUNCTION ( SQL SERVER )
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\n\n\nSELECT DISTINCT\n    event_day day,\n    emp_id,\n     SUM(out_time) OVER(PARTITION BY emp_id,event_day ) -  SUM(in_time) OVER(PARTITION BY emp_id,event_day )  total_time\nFROM\n    Employees\n\n```
['MySQL']