Minimum Time to Complete All Tasks
<p>There is a computer that can run an unlimited number of tasks <strong>at the same time</strong>. You are given a 2D integer array <code>tasks</code> where <code>tasks[i] = [start<sub>i</sub>, end<sub>i</sub>, duration<sub>i</sub>]</code> indicates that the <code>i<sup>th</sup></code> task should run for a total of <code>duration<sub>i</sub></code> seconds (not necessarily continuous) within the <strong>inclusive</strong> time range <code>[start<sub>i</sub>, end<sub>i</sub>]</code>.</p>

<p>You may turn on the computer only when it needs to run a task. You can also turn it off if it is idle.</p>

<p>Return <em>the minimum time during which the computer should be turned on to complete all tasks</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> tasks = [[2,3,1],[4,5,1],[1,5,2]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
- The first task can be run in the inclusive time range [2, 2].
- The second task can be run in the inclusive time range [5, 5].
- The third task can be run in the two inclusive time ranges [2, 2] and [5, 5].
The computer will be on for a total of 2 seconds.
</pre>

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

<pre>
<strong>Input:</strong> tasks = [[1,3,2],[2,5,3],[5,6,2]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 
- The first task can be run in the inclusive time range [2, 3].
- The second task can be run in the inclusive time ranges [2, 3] and [5, 5].
- The third task can be run in the two inclusive time range [5, 6].
The computer will be on for a total of 4 seconds.
</pre>

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

<ul>
	<li><code>1 &lt;= tasks.length &lt;= 2000</code></li>
	<li><code>tasks[i].length == 3</code></li>
	<li><code>1 &lt;= start<sub>i</sub>, end<sub>i</sub> &lt;= 2000</code></li>
	<li><code>1 &lt;= duration<sub>i</sub> &lt;= end<sub>i</sub> - start<sub>i</sub> + 1 </code></li>
</ul>
Stack; Greedy; Array; Binary Search; Sorting
function findMinimumTime(tasks: number[][]): number {
    tasks.sort((a, b) => a[1] - b[1]);
    const vis: number[] = Array(2010).fill(0);
    let ans = 0;
    for (let [start, end, duration] of tasks) {
        for (let i = start; i <= end; ++i) {
            duration -= vis[i];
        }
        for (let i = end; i >= start && duration > 0; --i) {
            if (vis[i] === 0) {
                --duration;
                ans += vis[i] = 1;
            }
        }
    }
    return ans;
}
