636. Exclusive Time of Functions

Posted 2020-12-19 johnnyzhao

package LeetCode_636

import java.util.*

/**
 * 636. Exclusive Time of Functions
 * https://leetcode.com/problems/exclusive-time-of-functions/description/
 *
 * On a single threaded CPU, we execute some functions.  Each function has a unique id between 0 and N-1.
We store logs in timestamp order that describe when a function is entered or exited.
Each log is a string with this format: "{function_id}:{"start" | "end"}:{timestamp}".
For example, "0:start:3" means the function with id 0 started at the beginning of timestamp 3.
"1:end:2" means the function with id 1 ended at the end of timestamp 2.
A function‘s exclusive time is the number of units of time spent in this function.
Note that this does not include any recursive calls to child functions.
The CPU is single threaded which means that only one function is being executed at a given time unit.
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]

Note:
1. 1 <= n <= 100
2. Two functions won‘t start or end at the same time.
3. Functions will always log when they exit.
 * */
class Solution {
    /*
    * solution: Stack,start to push in Stack, then end to pop from Stack
    * Time complexity:O(n), Space complexity:O(n/2), because do push and pop almost in same time
    * */
    fun exclusiveTime(n: Int, logs: List<String>): IntArray? {
        val stack = Stack<Int>()
        val result = IntArray(n)
        var prve = 0
        for (log in logs) {
            val list = log.split(":")
            val functionId = list.get(0).toInt()
            val operation = list.get(1)
            var time = list.get(2).toInt()
            if (operation == "start") {
                time -= 1
                //update the time
                if (stack.isNotEmpty()) {
                    result[stack.peek()] += time - prve
                }
                stack.push(functionId)
            } else {
                //like suspend it, pop it from stack
                result[stack.pop()] += time - prve
            }
            prve = time
        }
        return result
    }
}