<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:
  bool isNStraightHand(vector<int>& hand, int groupSize) {
    int n = hand.size();

    // Edge cases
    if (n % groupSize != 0) {
      return false;
    }
    if (groupSize == 1) {
      return true;
    }

    int k = n / groupSize;
    int done = 0;

    // Sort the hand
    sort(hand.begin(), hand.end());

    // Create a queue to store the pairs
    queue<pair<int, int>> q;
    pair<int, int> fr;

    // Loop through the sorted hand
    for (int i = 0; i < n; i++) {
      // Start of new straight group
      if (q.empty() || q.front().first + 1 != hand[i]) {
        q.push({hand[i], 1});
        continue;
      }

      // Continue the straight group
      fr = q.front();
      q.pop();
      if (fr.second != groupSize - 1) {
        q.push({hand[i], fr.second + 1});
      } else {
        // Finish the straight group
        done++;
      }
    }
    // Check if all groups are finished
    return (done == k);
  }
};
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