Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.9 MB (37.85%)

Author: jimit105

0266-palindrome-permutation/README.md

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+<p>Given a string <code>s</code>, return <code>true</code> <em>if a permutation of the string could form a </em><span data-keyword="palindrome-string"><em><strong>palindrome</strong></em></span><em> and </em><code>false</code><em> otherwise</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;code&quot;
+<strong>Output:</strong> false
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;aab&quot;
+<strong>Output:</strong> true
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;carerac&quot;
+<strong>Output:</strong> true
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 5000</code></li>
+	<li><code>s</code> consists of only lowercase English letters.</li>
+</ul>

0266-palindrome-permutation/solution.py

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+# Approach 2: HashMap
+
+# Time: O(n)
+# Space: O(n)
+
+from collections import Counter
+
+class Solution:
+    def canPermutePalindrome(self, s: str) -> bool:
+        count = Counter(s)
+        odds = sum(val % 2 for val in count.values())
+        return odds <= 1
+