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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 <= 1001 <= logs.length <= 5000 <= function_id < n0 <= 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
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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 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 for log in logs: 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; }