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636. Exclusive Time of Functions

Description

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.

Solutions

Solution 1: Stack + Simulation

We define a stack $\textit{stk}$ to store the identifiers of the currently executing functions. We also define an array $\textit{ans}$ to store the exclusive time of each function, initially setting the exclusive time of each function to $0$. We use a variable $\textit{pre}$ to record the previous timestamp.

We traverse the log array. For each log entry, we first split it by colons to get the function identifier $\textit{i}$, the operation type $\textit{op}$, and the timestamp $\textit{t}$.

If $\textit{op}$ is $\text{start}$, it means function $\textit{i}$ starts executing. We need to check if the stack is empty. If it is not empty, we add $\textit{cur} - \textit{pre}$ to the exclusive time of the function at the top of the stack, then push $\textit{i}$ onto the stack and update $\textit{pre}$ to $\textit{cur}$. If $\textit{op}$ is $\text{end}$, it means function $\textit{i}$ finishes executing. We add $\textit{cur} - \textit{pre} + 1$ to the exclusive time of the function at the top of the stack, then pop the top element from the stack and update $\textit{pre}$ to $\textit{cur} + 1$.

Finally, we return the array $\textit{ans}$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the log array.

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class Solution:
    def exclusiveTime(self, n: int, logs: List[str]) -> List[int]:
        stk = []
        ans = [0] * n
        pre = 0
        for log in logs:
            i, op, t = log.split(":")
            i, cur = int(i), int(t)
            if op[0] == "s":
                if stk:
                    ans[stk[-1]] += cur - pre
                stk.append(i)
                pre = cur
            else:
                ans[stk.pop()] += cur - pre + 1
                pre = cur + 1
        return ans

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class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        int[] ans = new int[n];
        Deque<Integer> stk = new ArrayDeque<>();
        int pre = 0;
        for (var log : logs) {
            var parts = log.split(":");
            int i = Integer.parseInt(parts[0]);
            int cur = Integer.parseInt(parts[2]);
            if (parts[1].charAt(0) == 's') {
                if (!stk.isEmpty()) {
                    ans[stk.peek()] += cur - pre;
                }
                stk.push(i);
                pre = cur;
            } else {
                ans[stk.pop()] += cur - pre + 1;
                pre = cur + 1;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> exclusiveTime(int n, vector<string>& logs) {
        vector<int> ans(n);
        stack<int> stk;
        int pre = 0;
        for (const auto& log : logs) {
            int i, cur;
            char c[10];
            sscanf(log.c_str(), "%d:%[^:]:%d", &i, c, &cur);
            if (c[0] == 's') {
                if (stk.size()) {
                    ans[stk.top()] += cur - pre;
                }
                stk.push(i);
                pre = cur;
            } else {
                ans[stk.top()] += cur - pre + 1;
                stk.pop();
                pre = cur + 1;
            }
        }
        return ans;
    }
};

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func exclusiveTime(n int, logs []string) []int {
    ans := make([]int, n)
    stk := []int{}
    pre := 0
    for _, log := range logs {
        parts := strings.Split(log, ":")
        i, _ := strconv.Atoi(parts[0])
        cur, _ := strconv.Atoi(parts[2])
        if parts[1][0] == 's' {
            if len(stk) > 0 {
                ans[stk[len(stk)-1]] += cur - pre
            }
            stk = append(stk, i)
            pre = cur
        } else {
            ans[stk[len(stk)-1]] += cur - pre + 1
            stk = stk[:len(stk)-1]
            pre = cur + 1
        }
    }
    return ans
}

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function exclusiveTime(n: number, logs: string[]): number[] {
    const ans: number[] = Array(n).fill(0);
    let pre = 0;
    const stk: number[] = [];
    for (const log of logs) {
        const [i, op, cur] = log.split(':');
        if (op[0] === 's') {
            if (stk.length) {
                ans[stk.at(-1)!] += +cur - pre;
            }
            stk.push(+i);
            pre = +cur;
        } else {
            ans[stk.pop()!] += +cur - pre + 1;
            pre = +cur + 1;
        }
    }
    return ans;
}

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