<p>Alice has some number of cards and she wants to rearrange the cards into groups so that each group is of size <code>groupSize</code>, and consists of <code>groupSize</code> consecutive cards.</p>

<p>Given an integer array <code>hand</code> where <code>hand[i]</code> is the value written on the <code>i<sup>th</sup></code> card and an integer <code>groupSize</code>, return <code>true</code> if she can rearrange the cards, or <code>false</code> otherwise.</p>

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

<pre>
<strong>Input:</strong> hand = [1,2,3,6,2,3,4,7,8], groupSize = 3
<strong>Output:</strong> true
<strong>Explanation:</strong> Alice&#39;s hand can be rearranged as [1,2,3],[2,3,4],[6,7,8]
</pre>

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

<pre>
<strong>Input:</strong> hand = [1,2,3,4,5], groupSize = 4
<strong>Output:</strong> false
<strong>Explanation:</strong> Alice&#39;s hand can not be rearranged into groups of 4.

</pre>

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

<ul>
	<li><code>1 &lt;= hand.length &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= hand[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= groupSize &lt;= hand.length</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Note:</strong> This question is the same as 1296: <a href="https://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/" target="_blank">https://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/</a></p>

class Solution {
public:
    typedef pair<int, int> pii;
    bool isNStraightHand(vector<int>& hand, int groupSize) {
        if (hand.size() % groupSize != 0) return false;
        unordered_map<int, int> mpp;
        // minimum queue
        priority_queue<pii, vector<pii>, greater<pii>> pq;

        // enter the frequencies
        for (int i = 0; i < hand.size(); i++) 
            mpp[hand[i]]++;

        // enter in the pq
        for (auto it = mpp.begin(); it != mpp.end(); it++) 
            pq.push({it->first, it->second});

        // fixing the logic
        while (!pq.empty()) {
            int top = pq.top().first;

            for (int j = top; j < top + groupSize; j++) {
                if (mpp.find(j) == mpp.end())
                    return false;
                mpp[j]--;
                if (mpp[j] == 0) {
                    if (j != pq.top().first) {
                        return false;
                    }
                    else {
                        mpp.erase(j);
                        pq.pop();
                    }
                }
            }    
        }
        return true;
    }
};
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