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# 0636. Exclusive Time of Functions

<https://leetcode.com/problems/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**](https://en.wikipedia.org/wiki/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:**

![](https://assets.leetcode.com/uploads/2019/04/05/diag1b.png)

```
**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 <= 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.

## ac: Stack

<https://leetcode.com/problems/exclusive-time-of-functions/discuss/153497/Java-solution-using-stack-wrapper-class-and-calculation-when-pop-element-from-the-stack>.

```java
class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        int[] functionTimes = new int[n];
        Deque<Log> stack = new ArrayDeque<>();
        
        for (String l : logs) {
            Log log = new Log(l);
            if (log.isStart) {
                stack.push(log);
            } else {
                Log top = stack.pop();
                functionTimes[top.functionId] += log.timestamp - top.timestamp + 1 - top.subDuration;
                if (!stack.isEmpty()) {
                    Log prev = stack.peek();
                    prev.subDuration += log.timestamp - top.timestamp + 1;
                }
            }
        }
        
        return functionTimes;
    }
    
    class Log {
        boolean isStart;
        int functionId;
        int timestamp;
        int subDuration = 0;
        
        Log(String log) {
            String[] parts = log.split(":");
            this.functionId = Integer.parseInt(parts[0]);
            this.isStart = "start".equals(parts[1]);
            this.timestamp = Integer.parseInt(parts[2]);
        }
    }
}
```

This is more ugly:

```java
class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        // edge cases
        if (n <= 0 || logs.size() == 0) return new int[0];

        Stack<int[]> stack = new Stack<>();
        int[] times = new int[n];


        for (String s : logs) {
            String[] info = s.split(":");
            int time = Integer.parseInt(info[2]);
            int func = Integer.parseInt(info[0]);

            if (stack.isEmpty()) {
                stack.push(new int[]{func, time});
            }

            if (info[1].equals("start")) {
                int[] prev = stack.peek();
                int gap = time - prev[1];
                times[prev[0]] += gap;
                stack.push(new int[]{func, time});
            } else { // end
                int gap = time - stack.pop()[1] + 1;
                times[func] += gap;
                stack.peek()[1] = time+1;
            }
        }

        return times;
    }
}

/*
stack, "start"/"end" have different behavior
*/
```