Exclusive Time of Functions
MediumUpdated: Aug 2, 2025
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Problem
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 acall 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 _theexclusive time of each function in an array, where the value at the _ith index represents the exclusive time for the function with IDi.
Examples
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 <= 1001 <= logs.length <= 5000 <= function_id < n0 <= timestamp <= 10^9- 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
Method 1 – Stack Simulation
Intuition
We use a stack to simulate the function call stack. For each log, we update the exclusive time for the function at the top of the stack. When a function starts, we push it onto the stack. When it ends, we pop it and update its exclusive time, making sure to account for nested calls.
Approach
- Initialize an array
ansof sizento store exclusive times. - Use a stack to keep track of the function call stack.
- Track the previous timestamp (
prev_time). - For each log:
- Parse the function id, type (start/end), and timestamp.
- If it's a start log:
- If the stack is not empty, add the time since
prev_timeto the function at the top of the stack. - Push the new function id onto the stack.
- Update
prev_timeto the current timestamp.
- If the stack is not empty, add the time since
- If it's an end log:
- Pop the function id from the stack.
- Add the time from
prev_timeto the current timestamp (inclusive) to its exclusive time. - Update
prev_timetotimestamp + 1.
- Return the
ansarray.
Code
C++
class Solution {
public:
vector<int> exclusiveTime(int n, vector<string>& logs) {
vector<int> ans(n, 0);
stack<int> st;
int prev = 0;
for (auto& log : logs) {
int p1 = log.find(':'), p2 = log.rfind(':');
int id = stoi(log.substr(0, p1));
string typ = log.substr(p1+1, p2-p1-1);
int t = stoi(log.substr(p2+1));
if (typ == "start") {
if (!st.empty()) ans[st.top()] += t - prev;
st.push(id);
prev = t;
} else {
ans[st.top()] += t - prev + 1;
st.pop();
prev = t + 1;
}
}
return ans;
}
};
Go
func exclusiveTime(n int, logs []string) []int {
ans := make([]int, n)
st := []int{}
prev := 0
for _, log := range logs {
parts := strings.Split(log, ":")
id, _ := strconv.Atoi(parts[0])
typ := parts[1]
t, _ := strconv.Atoi(parts[2])
if typ == "start" {
if len(st) > 0 {
ans[st[len(st)-1]] += t - prev
}
st = append(st, id)
prev = t
} else {
ans[st[len(st)-1]] += t - prev + 1
st = st[:len(st)-1]
prev = t + 1
}
}
return ans
}
Java
class Solution {
public int[] exclusiveTime(int n, List<String> logs) {
int[] ans = new int[n];
Deque<Integer> st = new ArrayDeque<>();
int prev = 0;
for (String log : logs) {
String[] parts = log.split(":");
int id = Integer.parseInt(parts[0]);
String typ = parts[1];
int t = Integer.parseInt(parts[2]);
if (typ.equals("start")) {
if (!st.isEmpty()) ans[st.peek()] += t - prev;
st.push(id);
prev = t;
} else {
ans[st.peek()] += t - prev + 1;
st.pop();
prev = t + 1;
}
}
return ans;
}
}
Kotlin
class Solution {
fun exclusiveTime(n: Int, logs: List<String>): IntArray {
val ans = IntArray(n)
val st = ArrayDeque<Int>()
var prev = 0
for (log in logs) {
val parts = log.split(":")
val id = parts[0].toInt()
val typ = parts[1]
val t = parts[2].toInt()
if (typ == "start") {
if (st.isNotEmpty()) ans[st.last()] += t - prev
st.addLast(id)
prev = t
} else {
ans[st.last()] += t - prev + 1
st.removeLast()
prev = t + 1
}
}
return ans
}
}
Python
class Solution:
def exclusiveTime(self, n: int, logs: list[str]) -> list[int]:
ans = [0] * n
st = []
prev = 0
for log in logs:
fid, typ, t = log.split(":")
fid, t = int(fid), int(t)
if typ == "start":
if st:
ans[st[-1]] += t - prev
st.append(fid)
prev = t
else:
ans[st[-1]] += t - prev + 1
st.pop()
prev = t + 1
return ans
Rust
impl Solution {
pub fn exclusive_time(n: i32, logs: Vec<String>) -> Vec<i32> {
let mut ans = vec![0; n as usize];
let mut st = vec![];
let mut prev = 0;
for log in logs {
let parts: Vec<&str> = log.split(':').collect();
let id = parts[0].parse::<usize>().unwrap();
let typ = parts[1];
let t = parts[2].parse::<i32>().unwrap();
if typ == "start" {
if let Some(&top) = st.last() {
ans[top] += t - prev;
}
st.push(id);
prev = t;
} else {
ans[*st.last().unwrap()] += t - prev + 1;
st.pop();
prev = t + 1;
}
}
ans
}
}
TypeScript
class Solution {
exclusiveTime(n: number, logs: string[]): number[] {
const ans = Array(n).fill(0);
const st: number[] = [];
let prev = 0;
for (const log of logs) {
const [fid, typ, t] = log.split(":");
const id = parseInt(fid), time = parseInt(t);
if (typ === "start") {
if (st.length) ans[st[st.length-1]] += time - prev;
st.push(id);
prev = time;
} else {
ans[st[st.length-1]] += time - prev + 1;
st.pop();
prev = time + 1;
}
}
return ans;
}
}
Complexity
- ⏰ Time complexity:
O(m), wheremis the number of logs. - 🧺 Space complexity:
O(n + m), for the answer array and stack.