636.Exclusive_Time_of_Functions(Medium).md

636. Exclusive Time of Functions (Medium)

Date and Time: Dec 31, 2024, 22:56 (EST)

Link: https://leetcode.com/problems/exclusive-time-of-functions

Question:

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.

Edge Case:

Input: n = 3, logs = ["0:start:0","1:start:2","2:start:3","2:end:4","1:end:5","0:end:6"]

Output: [3,2,2]

Edge Case:

Input: n = 3, logs = ["0:start:0","1:start:1","1:end:2","2:start:3","2:end:4","0:end:5"]

Output: [2,2,2]

Constraints:

• 1 <= n <= 100

• 1 <= logs.length <= 500

• 0 <= function_id < n

• 0 <= 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.

Walk-through:

1. Use stack[] to keep track of ids. And res[] to keep track of total execution times for each index i.

2. Split log into Id, status, pos. We first check if status == 'start', then check if stack[] is empty or not, if not, we add the difference pos - prevPos to previous index stack[-1], and append new index into stack[] and update prevPos = pos.
   If status == end, we update prev index stack.pop() by pos - prevPos + 1, then we update prevPos = pos + 1 (Otherwise it will increment (6-5+1 = 2) for res[1], instead of just 1).

Python Solution:

class Solution:
    def exclusiveTime(self, n: int, logs: List[str]) -> List[int]:
        # Save id into stack[], and res[i] ith func time
        # If status is 'start': curPos - prevPos to prev id. Then, update prevPos = curPos, append curId to stack
        # If status is 'end': pop to get the prev id, and update prev id by curPos-prevPos+1, and prevPos = curPos + 1

        # TC: O(n), n=len(logs), SC: O(n)
        stack, prevPos = [], 0
        res = [0] * n
        for log in logs:
            Id, status, pos = log.split(":")
            Id, pos = int(Id), int(pos)
            # For 'start', update the previous id just by pos - prevPos
            if status == 'start':
                # If it has previous start, update previous id
                if stack:
                    res[stack[-1]] += pos - prevPos
                stack.append(Id)
                prevPos = pos
            # For 'end', use current pos -  prevPos + 1
            else:
                res[stack.pop()] += pos - prevPos + 1
                prevPos = pos + 1
        return res

Time Complexity: $O(n)$
Space Complexity: $O(n)$