[leetcode] Exclusive Time of Functions

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.

 

 

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 units 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.

Example 4:

Input: n = 2, logs = ["0:start:0","0:start:2","0:end:5","1:start:7","1:end:7","0:end:8"]
Output: [8,1]

Example 5:

Input: n = 1, logs = ["0:start:0","0:end:0"]
Output: [1]

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.

single-thread CPU 가 프로그램을 스케쥴링하는 방식을 모방한 문제이다. CPU 가 function 을 실행할때마다 call stack 에 function 의 정보들이 저장되게 되며, 순차적으로 실행될때마다 스택에서 제외되게 된다.

 

이때 각 function 의 ID 마다 실행되는 시간을 추적하여 저장하게끔 되어있다. 단 주의해야할 점은 recursive call 이 일어나는 경우인데, recursive call 이 같은 function 에서 일어나는 경우 같은 ID 의 수행시간에 더해지게 된다.

 

나는 이 문제를 stack 을 이용해서 풀었는데, 각 pair 마다 start 시간과, 하위 child 의 수행 시간을 가지고 있게 된다. 그러면 stack 의 제일 위에 있는 function 의 수행시간은 호출이 끝난 시간 - 호출이 시작한 시간 - child function 의 수행시간 이 된다.

 

class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        int [] res = new int[n];

        Stack<Pair> stack = new Stack<>();

        for (String log : logs) {
            String [] split = log.split(":");

            int id = Integer.valueOf(split[0]);
            String op = split[1];
            int time = Integer.valueOf(split[2]);

            if (op.equals("start")) {
                stack.add(new Pair(time, 0));
            } else {
                Pair pop = stack.pop();

                int executeTime = time - pop.start + 1 - pop.child;

                res[id] += executeTime;

                if (stack.isEmpty() == false) {
                    Pair correctPair = stack.pop();

                    correctPair.child += executeTime + pop.child;

                    stack.add(correctPair);
                }
            }
        }

        return res;
    }

  static class Pair {
        int start;
        int child;

        public Pair(int start, int child) {
            this.start = start;
            this.child = child;
        }
    }
}