# 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

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

• class Solution {
public:
vector<int> exclusiveTime(int n, vector<string>& logs) {
vector<int> ans(n);
stack<int> stk;
int curr = -1;
for (auto& log : logs) {
char type[10];
int fid, ts;
sscanf(log.c_str(), "%d:%[^:]:%d", &fid, type, &ts);
if (type[0] == 's') {
if (!stk.empty()) ans[stk.top()] += ts - curr;
curr = ts;
stk.push(fid);
} else {
fid = stk.top();
stk.pop();
ans[fid] += ts - curr + 1;
curr = ts + 1;
}
}
return ans;
}
};

• class Solution:
def exclusiveTime(self, n: int, logs: List[str]) -> List[int]:
ans = [0] * n
stk = []
curr = -1
t = log.split(':')
fid = int(t[0])
ts = int(t[2])
if t[1] == 'start':
if stk:
ans[stk[-1]] += ts - curr
stk.append(fid)
curr = ts
else:
fid = stk.pop()
ans[fid] += ts - curr + 1
curr = ts + 1
return ans


• func exclusiveTime(n int, logs []string) []int {
ans := make([]int, n)
stk := []int{}
curr := 1
for _, log := range logs {
t := strings.Split(log, ":")
fid, _ := strconv.Atoi(t[0])
ts, _ := strconv.Atoi(t[2])
if t[1][0] == 's' {
if len(stk) > 0 {
ans[stk[len(stk)-1]] += ts - curr
}
stk = append(stk, fid)
curr = ts
} else {
fid := stk[len(stk)-1]
stk = stk[:len(stk)-1]
ans[fid] += ts - curr + 1
curr = ts + 1
}
}
return ans
}

• function exclusiveTime(n: number, logs: string[]): number[] {
const res = new Array(n).fill(0);
const stack: [number, number][] = [];

for (const log of logs) {
const t = log.split(':');
const [id, state, time] = [Number(t[0]), t[1], Number(t[2])];

if (state === 'start') {
if (stack.length !== 0) {
const pre = stack[stack.length - 1];
res[pre[0]] += time - pre[1];
}
stack.push([id, time]);
} else {
const pre = stack.pop();
res[pre[0]] += time - pre[1] + 1;
if (stack.length !== 0) {
stack[stack.length - 1][1] = time + 1;
}
}
}

return res;
}