[LeetCode Sync] Runtime - 3 ms (42.52%), Memory - 18 MB (46.41%)

Author: NinePiece2

Solutions/0889-buddy-strings/README.md

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+<p>Given two strings <code>s</code> and <code>goal</code>, return <code>true</code><em> if you can swap two letters in </em><code>s</code><em> so the result is equal to </em><code>goal</code><em>, otherwise, return </em><code>false</code><em>.</em></p>
+
+<p>Swapping letters is defined as taking two indices <code>i</code> and <code>j</code> (0-indexed) such that <code>i != j</code> and swapping the characters at <code>s[i]</code> and <code>s[j]</code>.</p>
+
+<ul>
+	<li>For example, swapping at indices <code>0</code> and <code>2</code> in <code>&quot;abcd&quot;</code> results in <code>&quot;cbad&quot;</code>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;ab&quot;, goal = &quot;ba&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> You can swap s[0] = &#39;a&#39; and s[1] = &#39;b&#39; to get &quot;ba&quot;, which is equal to goal.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;ab&quot;, goal = &quot;ab&quot;
+<strong>Output:</strong> false
+<strong>Explanation:</strong> The only letters you can swap are s[0] = &#39;a&#39; and s[1] = &#39;b&#39;, which results in &quot;ba&quot; != goal.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;aa&quot;, goal = &quot;aa&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> You can swap s[0] = &#39;a&#39; and s[1] = &#39;a&#39; to get &quot;aa&quot;, which is equal to goal.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length, goal.length &lt;= 2 * 10<sup>4</sup></code></li>
+	<li><code>s</code> and <code>goal</code> consist of lowercase letters.</li>
+</ul>

Solutions/0889-buddy-strings/solution.py

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+class Solution:
+    def buddyStrings(self, s: str, goal: str) -> bool:
+        m, n = len(s), len(goal)
+        if m != n:
+            return False
+
+        count_s, count_goal = Counter(s), Counter(goal)
+        if count_s != count_goal:
+            return False
+        
+        diff = sum(s[i] != goal[i] for i in range(n))
+        return diff == 2 or (diff == 0 and any(val > 1 for val in count_s.values()))