LeetCode in Kotlin

636. Exclusive Time of Functions

Medium

On a single-threaded CPU, we execute a program containing n functions. Each function has a unique ID between 0 and n-1.

Function calls are stored in a call stack: when a function call starts, its ID is pushed onto the stack, and when a function call ends, its ID is popped off the stack. The function whose ID is at the top of the stack is the current function being executed. Each time a function starts or ends, we write a log with the ID, whether it started or ended, and the timestamp.

You are given a list logs, where logs[i] represents the ith log message formatted as a string "{function_id}:{"start" | "end"}:{timestamp}". For example, "0:start:3" means a function call with function ID 0 started at the beginning of timestamp 3, and "1:end:2" means a function call with function ID 1 ended at the end of timestamp 2. Note that a function can be called multiple times, possibly recursively.

A function’s exclusive time is the sum of execution times for all function calls in the program. For example, if a function is called twice, one call executing for 2 time units and another call executing for 1 time unit, the exclusive time is 2 + 1 = 3.

Return the exclusive time of each function in an array, where the value at the ith index represents the exclusive time for the function with ID i.

Example 1:

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 for units of time and reaches the end of time 1.

Function 1 starts at the beginning of time 2, executes for 4 units of time, and ends at the end of time 5.

Function 0 resumes execution at the beginning of time 6 and executes 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.

Example 2:

Input: n = 1, logs = [“0:start:0”,”0:start:2”,”0:end:5”,”0:start:6”,”0:end:6”,”0:end:7”]

Output: [8]

Explanation:

Function 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself.

Function 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time.

Function 0 (initial call) resumes execution then immediately calls itself again.

Function 0 (2nd recursive call) starts at the beginning of time 6 and executes for 1 unit of time.

Function 0 (initial call) resumes execution at the beginning of time 7 and executes for 1 unit of time.

So function 0 spends 2 + 4 + 1 + 1 = 8 units of total time executing.

Example 3:

Input: n = 2, logs = [“0:start:0”,”0:start:2”,”0:end:5”,”1:start:6”,”1:end:6”,”0:end:7”]

Output: [7,1]

Explanation:

Function 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself.

Function 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time.

Function 0 (initial call) resumes execution then immediately calls function 1.

Function 1 starts at the beginning of time 6, executes 1 unit of time, and ends at the end of time 6.

Function 0 resumes execution at the beginning of time 6 and executes for 2 units of time.

So function 0 spends 2 + 4 + 1 = 7 units of total time executing, and function 1 spends 1 unit of total time executing.

Constraints:

• 1 <= n <= 100
• 1 <= logs.length <= 500
• 0 <= function_id < n
• 0 <= timestamp <= 109
• No two start events will happen at the same timestamp.
• No two end events will happen at the same timestamp.
• Each function has an "end" log for each "start" log.

Solution

import java.util.ArrayDeque
import java.util.Deque

class Solution {
    fun exclusiveTime(n: Int, logs: List<String>): IntArray {
        val stack: Deque<Log> = ArrayDeque()
        val result = IntArray(n)
        for (content in logs) {
            val log = Log(content)
            if (log.isStart) {
                stack.push(log)
            } else {
                val top = stack.pop()
                val executionTime = log.time - top.time + 1
                result[top.id] += executionTime - top.waitingTime
                if (stack.isNotEmpty()) {
                    stack.peek().waitingTime += executionTime
                }
            }
        }
        return result
    }

    private class Log internal constructor(content: String) {
        var id: Int
        var isStart: Boolean
        var time: Int
        var waitingTime: Int
        init {
            val tokens = content.split(":".toRegex()).dropLastWhile { it.isEmpty() }.toTypedArray()
            id = tokens[0].toInt()
            isStart = tokens[1] == "start"
            time = tokens[2].toInt()
            waitingTime = 0
        }
    }
}

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