GoLang Solution For LeetCode Problem: Exclusive Time of Functions
Solving Exclusive Time of Functions in go. Please try yourself first to solve the problem and submit your implementation to LeetCode before looking into solution.
We store logs in timestamp order that describe when a function is entered or exited.
Return the exclusive time of each function, sorted by their function id.
Input: n = 2 logs = ["0:start:0","1:start:2","1:end:5","0:end:6"] Output: [3, 4] Explanation: Function 0 starts at the beginning of time 0, then it executes 2 units of time and reaches the end of time 1. Now function 1 starts at the beginning of time 2, executes 4 units of time and ends at time 5. Function 0 is running again at the beginning of time 6, and also ends at the end of time 6, thus executing for 1 unit of time. So function 0 spends 2 + 1 = 3 units of total time executing, and function 1 spends 4 units of total time executing.
1 <= n <= 100
- Two functions won't start or end at the same time.
- Functions will always log when they exit.
See the full details of the problem Exclusive Time of Functions at LeetCode
Originally posted at: @github.com/halfrost/LeetCode-Go