Minimum Time to Repair Cars
<p>You are given an integer array <code>ranks</code> representing the <strong>ranks</strong> of some mechanics. <font face="monospace">ranks<sub>i</sub></font> is the rank of the <font face="monospace">i<sup>th</sup></font> mechanic<font face="monospace">.</font> A mechanic with a rank <code>r</code> can repair <font face="monospace">n</font> cars in <code>r * n<sup>2</sup></code> minutes.</p>

<p>You are also given an integer <code>cars</code> representing the total number of cars waiting in the garage to be repaired.</p>

<p>Return <em>the <strong>minimum</strong> time taken to repair all the cars.</em></p>

<p><strong>Note:</strong> All the mechanics can repair the cars simultaneously.</p>

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

<pre>
<strong>Input:</strong> ranks = [4,2,3,1], cars = 10
<strong>Output:</strong> 16
<strong>Explanation:</strong> 
- The first mechanic will repair two cars. The time required is 4 * 2 * 2 = 16 minutes.
- The second mechanic will repair two cars. The time required is 2 * 2 * 2 = 8 minutes.
- The third mechanic will repair two cars. The time required is 3 * 2 * 2 = 12 minutes.
- The fourth mechanic will repair four cars. The time required is 1 * 4 * 4 = 16 minutes.
It can be proved that the cars cannot be repaired in less than 16 minutes.​​​​​
</pre>

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

<pre>
<strong>Input:</strong> ranks = [5,1,8], cars = 6
<strong>Output:</strong> 16
<strong>Explanation:</strong> 
- The first mechanic will repair one car. The time required is 5 * 1 * 1 = 5 minutes.
- The second mechanic will repair four cars. The time required is 1 * 4 * 4 = 16 minutes.
- The third mechanic will repair one car. The time required is 8 * 1 * 1 = 8 minutes.
It can be proved that the cars cannot be repaired in less than 16 minutes.​​​​​
</pre>

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

<ul>
	<li><code>1 &lt;= ranks.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= ranks[i] &lt;= 100</code></li>
	<li><code>1 &lt;= cars &lt;= 10<sup>6</sup></code></li>
</ul>
Array; Binary Search
class Solution {
public:
    long long repairCars(vector<int>& ranks, int cars) {
        long long left = 0, right = 1LL * ranks[0] * cars * cars;
        while (left < right) {
            long long mid = (left + right) >> 1;
            long long cnt = 0;
            for (int r : ranks) {
                cnt += sqrt(mid / r);
            }
            if (cnt >= cars) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};