Task Scheduler
<p>You are given an array of CPU <code>tasks</code>, each labeled with a letter from A to Z, and a number <code>n</code>. Each CPU interval can be idle or allow the completion of one task. Tasks can be completed in any order, but there&#39;s a constraint: there has to be a gap of <strong>at least</strong> <code>n</code> intervals between two tasks with the same label.</p>

<p>Return the <strong>minimum</strong> number of CPU intervals required to complete all tasks.</p>

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

<div class="example-block" style="
    border-color: var(--border-tertiary);
    border-left-width: 2px;
    color: var(--text-secondary);
    font-size: .875rem;
    margin-bottom: 1rem;
    margin-top: 1rem;
    overflow: visible;
    padding-left: 1rem;
">
<p><strong>Input:</strong> <span class="example-io" style="
    font-family: Menlo,sans-serif;
    font-size: 0.85rem;
">tasks = [&quot;A&quot;,&quot;A&quot;,&quot;A&quot;,&quot;B&quot;,&quot;B&quot;,&quot;B&quot;], n = 2</span></p>

<p><strong>Output:</strong> <span class="example-io" style="
font-family: Menlo,sans-serif;
font-size: 0.85rem;
">8</span></p>

<p><strong>Explanation:</strong> A possible sequence is: A -&gt; B -&gt; idle -&gt; A -&gt; B -&gt; idle -&gt; A -&gt; B.</p>

<p>After completing task A, you must wait two intervals before doing A again. The same applies to task B. In the 3<sup>rd</sup> interval, neither A nor B can be done, so you idle. By the 4<sup>th</sup> interval, you can do A again as 2 intervals have passed.</p>
</div>

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

<div class="example-block" style="
    border-color: var(--border-tertiary);
    border-left-width: 2px;
    color: var(--text-secondary);
    font-size: .875rem;
    margin-bottom: 1rem;
    margin-top: 1rem;
    overflow: visible;
    padding-left: 1rem;
">
<p><strong>Input:</strong> <span class="example-io" style="
    font-family: Menlo,sans-serif;
    font-size: 0.85rem;
">tasks = [&quot;A&quot;,&quot;C&quot;,&quot;A&quot;,&quot;B&quot;,&quot;D&quot;,&quot;B&quot;], n = 1</span></p>

<p><strong>Output:</strong> <span class="example-io" style="
    font-family: Menlo,sans-serif;
    font-size: 0.85rem;
">6</span></p>

<p><strong>Explanation:</strong> A possible sequence is: A -&gt; B -&gt; C -&gt; D -&gt; A -&gt; B.</p>

<p>With a cooling interval of 1, you can repeat a task after just one other task.</p>
</div>

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

<div class="example-block" style="
    border-color: var(--border-tertiary);
    border-left-width: 2px;
    color: var(--text-secondary);
    font-size: .875rem;
    margin-bottom: 1rem;
    margin-top: 1rem;
    overflow: visible;
    padding-left: 1rem;
">
<p><strong>Input:</strong> <span class="example-io" style="
    font-family: Menlo,sans-serif;
    font-size: 0.85rem;
">tasks = [&quot;A&quot;,&quot;A&quot;,&quot;A&quot;, &quot;B&quot;,&quot;B&quot;,&quot;B&quot;], n = 3</span></p>

<p><strong>Output:</strong> <span class="example-io" style="
    font-family: Menlo,sans-serif;
    font-size: 0.85rem;
">10</span></p>

<p><strong>Explanation:</strong> A possible sequence is: A -&gt; B -&gt; idle -&gt; idle -&gt; A -&gt; B -&gt; idle -&gt; idle -&gt; A -&gt; B.</p>

<p>There are only two types of tasks, A and B, which need to be separated by 3 intervals. This leads to idling twice between repetitions of these tasks.</p>
</div>

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

<ul>
	<li><code>1 &lt;= tasks.length &lt;= 10<sup>4</sup></code></li>
	<li><code>tasks[i]</code> is an uppercase English letter.</li>
	<li><code>0 &lt;= n &lt;= 100</code></li>
</ul>
Greedy; Array; Hash Table; Counting; Sorting; Heap (Priority Queue)
class Solution {
public:
    int leastInterval(vector<char>& tasks, int n) {
        vector<int> cnt(26);
        int x = 0;
        for (char c : tasks) {
            c -= 'A';
            ++cnt[c];
            x = max(x, cnt[c]);
        }
        int s = 0;
        for (int v : cnt) {
            s += v == x;
        }
        return max((int) tasks.size(), (x - 1) * (n + 1) + s);
    }
};