Exclusive Time of Functions
<p>On a <strong>single-threaded</strong> CPU, we execute a program containing <code>n</code> functions. Each function has a unique ID between <code>0</code> and <code>n-1</code>.</p>

<p>Function calls are <strong>stored in a <a href="https://en.wikipedia.org/wiki/Call_stack">call stack</a></strong>: 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 <strong>the current function being executed</strong>. Each time a function starts or ends, we write a log with the ID, whether it started or ended, and the timestamp.</p>

<p>You are given a list <code>logs</code>, where <code>logs[i]</code> represents the <code>i<sup>th</sup></code> log message formatted as a string <code>&quot;{function_id}:{&quot;start&quot; | &quot;end&quot;}:{timestamp}&quot;</code>. For example, <code>&quot;0:start:3&quot;</code> means a function call with function ID <code>0</code> <strong>started at the beginning</strong> of timestamp <code>3</code>, and <code>&quot;1:end:2&quot;</code> means a function call with function ID <code>1</code> <strong>ended at the end</strong> of timestamp <code>2</code>. Note that a function can be called <b>multiple times, possibly recursively</b>.</p>

<p>A function&#39;s <strong>exclusive time</strong> 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 <code>2</code> time units and another call executing for <code>1</code> time unit, the <strong>exclusive time</strong> is <code>2 + 1 = 3</code>.</p>

<p>Return <em>the <strong>exclusive time</strong> of each function in an array, where the value at the </em><code>i<sup>th</sup></code><em> index represents the exclusive time for the function with ID </em><code>i</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0636.Exclusive%20Time%20of%20Functions/images/diag1b.png" style="width: 550px; height: 239px;" />
<pre>
<strong>Input:</strong> n = 2, logs = [&quot;0:start:0&quot;,&quot;1:start:2&quot;,&quot;1:end:5&quot;,&quot;0:end:6&quot;]
<strong>Output:</strong> [3,4]
<strong>Explanation:</strong>
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1, logs = [&quot;0:start:0&quot;,&quot;0:start:2&quot;,&quot;0:end:5&quot;,&quot;0:start:6&quot;,&quot;0:end:6&quot;,&quot;0:end:7&quot;]
<strong>Output:</strong> [8]
<strong>Explanation:</strong>
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.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 2, logs = [&quot;0:start:0&quot;,&quot;0:start:2&quot;,&quot;0:end:5&quot;,&quot;1:start:6&quot;,&quot;1:end:6&quot;,&quot;0:end:7&quot;]
<strong>Output:</strong> [7,1]
<strong>Explanation:</strong>
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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>2 &lt;= logs.length &lt;= 500</code></li>
	<li><code>0 &lt;= function_id &lt; n</code></li>
	<li><code>0 &lt;= timestamp &lt;= 10<sup>9</sup></code></li>
	<li>No two start events will happen at the same timestamp.</li>
	<li>No two end events will happen at the same timestamp.</li>
	<li>Each function has an <code>&quot;end&quot;</code> log for each <code>&quot;start&quot;</code> log.</li>
</ul>
Stack; Array
function exclusiveTime(n: number, logs: string[]): number[] {
    const ans: number[] = Array(n).fill(0);
    let pre = 0;
    const stk: number[] = [];
    for (const log of logs) {
        const [i, op, cur] = log.split(':');
        if (op[0] === 's') {
            if (stk.length) {
                ans[stk.at(-1)!] += +cur - pre;
            }
            stk.push(+i);
            pre = +cur;
        } else {
            ans[stk.pop()!] += +cur - pre + 1;
            pre = +cur + 1;
        }
    }
    return ans;
}
