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.

def exclusiveTime(n: int, logs: list[str]) -> list[int]:
    ans = [0] * n
    stk = []
    pre = 0
    for log in logs:
        i, op, t = log.split(':')
        i, t = int(i), int(t)
        if op == 'start':
            if stk:
                ans[stk[-1]] += t - pre
            stk.append(i)
            pre = t
        else:
            ans[stk.pop()] += t - pre + 1
            pre = t + 1
    return ans
 

import java.util.*;
 
class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        int[] ans = new int[n];
        Deque<Integer> stk = new ArrayDeque<>();
        int pre = 0;
        for (String log : logs) {
            String[] parts = log.split(":");
            int i = Integer.parseInt(parts[0]);
            int t = Integer.parseInt(parts[2]);
            String op = parts[1];
            if (op.equals("start")) {
                if (!stk.isEmpty()) {
                    ans[stk.peek()] += t - pre;
                }
                stk.push(i);
                pre = t;
            } else {
                ans[stk.pop()] += t - pre + 1;
                pre = t + 1;
            }
        }
        return ans;
    }
}