Sync LeetCode submission Runtime - 51 ms (25.11%), Memory - 18.2 MB (8.81%)

Author: jimit105

1502-construct-k-palindrome-strings/README.md

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+<p>Given a string <code>s</code> and an integer <code>k</code>, return <code>true</code> <em>if you can use all the characters in </em><code>s</code><em> to construct </em><code>k</code><em> palindrome strings or </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;annabelle&quot;, k = 2
+<strong>Output:</strong> true
+<strong>Explanation:</strong> You can construct two palindromes using all characters in s.
+Some possible constructions &quot;anna&quot; + &quot;elble&quot;, &quot;anbna&quot; + &quot;elle&quot;, &quot;anellena&quot; + &quot;b&quot;
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;leetcode&quot;, k = 3
+<strong>Output:</strong> false
+<strong>Explanation:</strong> It is impossible to construct 3 palindromes using all the characters of s.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;true&quot;, k = 4
+<strong>Output:</strong> true
+<strong>Explanation:</strong> The only possible solution is to put each character in a separate string.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s</code> consists of lowercase English letters.</li>
+	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
+</ul>

1502-construct-k-palindrome-strings/solution.py

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+# Approach 1: Count Odd Frequencies
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def canConstruct(self, s: str, k: int) -> bool:
+        if len(s) < k:
+            return False
+        if len(s) == k:
+            return True
+
+        freq = [0] * 26
+        odd_count = 0
+
+        for char in s:
+            freq[ord(char) - ord('a')] += 1
+
+        for count in freq:
+            if count % 2 == 1:
+                odd_count += 1
+
+        return odd_count <= k
+