Minimum Number of Seconds to Make Mountain Height Zero
<p>You are given an integer <code>mountainHeight</code> denoting the height of a mountain.</p>

<p>You are also given an integer array <code>workerTimes</code> representing the work time of workers in <strong>seconds</strong>.</p>

<p>The workers work <strong>simultaneously</strong> to <strong>reduce</strong> the height of the mountain. For worker <code>i</code>:</p>

<ul>
	<li>To decrease the mountain&#39;s height by <code>x</code>, it takes <code>workerTimes[i] + workerTimes[i] * 2 + ... + workerTimes[i] * x</code> seconds. For example:

    <ul>
    	<li>To reduce the height of the mountain by 1, it takes <code>workerTimes[i]</code> seconds.</li>
    	<li>To reduce the height of the mountain by 2, it takes <code>workerTimes[i] + workerTimes[i] * 2</code> seconds, and so on.</li>
    </ul>
    </li>

</ul>

<p>Return an integer representing the <strong>minimum</strong> number of seconds required for the workers to make the height of the mountain 0.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">mountainHeight = 4, workerTimes = [2,1,1]</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>One way the height of the mountain can be reduced to 0 is:</p>

<ul>
	<li>Worker 0 reduces the height by 1, taking <code>workerTimes[0] = 2</code> seconds.</li>
	<li>Worker 1 reduces the height by 2, taking <code>workerTimes[1] + workerTimes[1] * 2 = 3</code> seconds.</li>
	<li>Worker 2 reduces the height by 1, taking <code>workerTimes[2] = 1</code> second.</li>
</ul>

<p>Since they work simultaneously, the minimum time needed is <code>max(2, 3, 1) = 3</code> seconds.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">mountainHeight = 10, workerTimes = [3,2,2,4]</span></p>

<p><strong>Output:</strong> <span class="example-io">12</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li>Worker 0 reduces the height by 2, taking <code>workerTimes[0] + workerTimes[0] * 2 = 9</code> seconds.</li>
	<li>Worker 1 reduces the height by 3, taking <code>workerTimes[1] + workerTimes[1] * 2 + workerTimes[1] * 3 = 12</code> seconds.</li>
	<li>Worker 2 reduces the height by 3, taking <code>workerTimes[2] + workerTimes[2] * 2 + workerTimes[2] * 3 = 12</code> seconds.</li>
	<li>Worker 3 reduces the height by 2, taking <code>workerTimes[3] + workerTimes[3] * 2 = 12</code> seconds.</li>
</ul>

<p>The number of seconds needed is <code>max(9, 12, 12, 12) = 12</code> seconds.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">mountainHeight = 5, workerTimes = [1]</span></p>

<p><strong>Output:</strong> <span class="example-io">15</span></p>

<p><strong>Explanation:</strong></p>

<p>There is only one worker in this example, so the answer is <code>workerTimes[0] + workerTimes[0] * 2 + workerTimes[0] * 3 + workerTimes[0] * 4 + workerTimes[0] * 5 = 15</code>.</p>
</div>

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

<ul>
	<li><code>1 &lt;= mountainHeight &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= workerTimes.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= workerTimes[i] &lt;= 10<sup>6</sup></code></li>
</ul>
Greedy; Array; Math; Binary Search; Heap (Priority Queue)
function minNumberOfSeconds(mountainHeight: number, workerTimes: number[]): number {
    const check = (t: bigint): boolean => {
        let h = BigInt(0);
        for (const wt of workerTimes) {
            h += BigInt(Math.floor(Math.sqrt((Number(t) * 2.0) / wt + 0.25) - 0.5));
        }
        return h >= BigInt(mountainHeight);
    };

    let l = BigInt(1);
    let r = BigInt(1e16);

    while (l < r) {
        const mid = (l + r) >> BigInt(1);
        if (check(mid)) {
            r = mid;
        } else {
            l = mid + 1n;
        }
    }

    return Number(l);
}
