Sync LeetCode submission Runtime - 11 ms (74.27%), Memory - 18 MB (25.35%)

Author: jimit105

0242-valid-anagram/README.md

@@ -1,15 +1,22 @@
-<p>Given two strings <code>s</code> and <code>t</code>, return <code>true</code> <em>if</em> <code>t</code> <em>is an anagram of</em> <code>s</code><em>, and</em> <code>false</code> <em>otherwise</em>.</p>
-
-<p>An <strong>Anagram</strong> is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.</p>
+<p>Given two strings <code>s</code> and <code>t</code>, return <code>true</code> if <code>t</code> is an <span data-keyword="anagram">anagram</span> of <code>s</code>, and <code>false</code> otherwise.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
-<pre><strong>Input:</strong> s = "anagram", t = "nagaram"
-<strong>Output:</strong> true
-</pre><p><strong class="example">Example 2:</strong></p>
-<pre><strong>Input:</strong> s = "rat", t = "car"
-<strong>Output:</strong> false
-</pre>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;anagram&quot;, t = &quot;nagaram&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">true</span></p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;rat&quot;, t = &quot;car&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">false</span></p>
+</div>
+
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 

0242-valid-anagram/solution.py

@@ -1,8 +1,18 @@
-from collections import Counter
+# Approach 2: Frequency Counter
+
+# Time: O(n)
+# Space: O(1)
+
 class Solution:
     def isAnagram(self, s: str, t: str) -> bool:
-        s_count = Counter(s)
-        t_count = Counter(t)
-        
-        return True if s_count == t_count else False
+        if len(s) != len(t):
+            return False
+
+        count = [0] * 26
+
+        for i in range(len(s)):
+            count[ord(s[i]) - ord('a')] += 1
+            count[ord(t[i]) - ord('a')] -= 1
+
+        return all(c == 0 for c in count)