<p>You are given an array of CPU <code>tasks</code>, each represented by letters&nbsp;A&nbsp;to Z, and a cooling time, <code>n</code>. Each cycle or interval allows the completion of one task. Tasks can be completed in any order, but there&#39;s a constraint: <strong>identical</strong> tasks must be separated by at least <code>n</code> intervals due to cooling time.</p>

<p>​Return the <em>minimum number of intervals</em> 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 cycles 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> cycle, 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>

class Solution {
public:
    int leastInterval(vector<char>& tasks, int n) {
        int size=tasks.size();
        priority_queue<pair<int,char>> pq;
        unordered_map<char,int> mp;
        vector<char> ans(size+(size*n)+1,'1');

        for(int i=0;i<size;i++){
            mp[tasks[i]]++;
        }
        for(auto it:mp){
            pq.push({it.second,it.first});
        }
        int start=0;
        while(!pq.empty()){
            int freq=pq.top().first;
            int ch=pq.top().second;

            pq.pop();
            int i=start;
            while(freq--){
                if(ans[i]=='1'){
                    ans[i]=ch;
                }
                else{
                    for(int j=i+1;i<size+(size*n);j++){
                        if(ans[j]=='1'){
                    ans[j]=ch;
                    break;
                }
                    }
                }
                
                i+=(n+1);
            }
            start++;
        }
        for(int i=size+(size*n);i>=0;i--){
            if(ans[i]!='1'){
                return i+1;
            }
        }
        return 0;
    }
};
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