MySQL Solution for Find Total Time Spent by Each Employee Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the solution is to use the GROUP BY clause to group the rows of th
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the solution is to use the GROUP BY clause to group the rows of the Employees table based on the event_day and emp_id columns. This allows us to calculate the total time spent by each employee on each day.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe query uses the GROUP BY clause to group the rows of the Employees table based on the event_day and emp_id columns.\n\nWithin each group, we use the SUM(out_time - in_time) expression to calculate the total time spent by each employee on each day. This expression calculates the difference between the out_time and in_time for each row within the group and then sums up the differences.\n\nThe result is a table with the day, emp_id, and total_time columns, representing the day, employee ID, and total time spent by each employee on each day at the office.\n\nThe query returns the result table in any order.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of the solution depends on the number of rows in the Employees table. If there are n rows in the table, the query will require traversing each row once to perform the grouping and calculate the total time for each group. Therefore, the time complexity is O(n).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of the solution depends on the number of distinct combinations of event_day and emp_id in the Employees table. The query will store the distinct combinations along with the corresponding total time in memory. If there are m distinct combinations, the space complexity is O(m)\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT event_day AS day, emp_id, SUM(out_time - in_time) AS total_time\nFROM Employees\nGROUP BY day, emp_id;\n\n```
['MySQL']