Sync LeetCode submission Runtime - 157 ms (75.82%), Memory - 18.5 MB (16.56%)

Author: jimit105

0876-hand-of-straights/README.md

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+<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>

0876-hand-of-straights/solution.py

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+# Approach 1: Using Map
+
+from collections import Counter
+import heapq
+
+class Solution:
+    def isNStraightHand(self, hand: List[int], groupSize: int) -> bool:
+        hand_size = len(hand)
+
+        if hand_size % groupSize != 0:
+            return False
+
+        card_count = Counter(hand)
+
+        # Min-heap to process the cards in sorted order
+        min_heap = list(card_count.keys())
+        heapq.heapify(min_heap)
+        
+        while min_heap:
+            current_card = min_heap[0] # Get smallest card
+
+            for i in range(groupSize):
+                if card_count[current_card + i] == 0:
+                    return False
+                card_count[current_card + i] -= 1
+                if card_count[current_card + i] == 0:
+                    if current_card + i != heapq.heappop(min_heap):
+                        return False
+
+        return True