<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:
    def isNStraightHand(self, hand: List[int], groupSize: int) -> bool:
        if len(hand)%groupSize!=0:
            return False
        h={}
        hand=sorted(hand)
        for n in hand:
            if not n in h:
                h[n]=0
            h[n]+=1
        for i,n in enumerate(hand):
            if h[n]>0:
                j=1
                h[n]-=1
                while j<groupSize:
                    if n+j in h and h[n+j]>0:
                        h[n+j]-=1
                    else:
                        return False
                    j+=1
        return sum(h.values())==0

                        
                         
                

       
18500