Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.8 MB (47.15%)

Author: jimit105

1915-check-if-one-string-swap-can-make-strings-equal/README.md

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+<p>You are given two strings <code>s1</code> and <code>s2</code> of equal length. A <strong>string swap</strong> is an operation where you choose two indices in a string (not necessarily different) and swap the characters at these indices.</p>
+
+<p>Return <code>true</code> <em>if it is possible to make both strings equal by performing <strong>at most one string swap </strong>on <strong>exactly one</strong> of the strings. </em>Otherwise, return <code>false</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s1 = &quot;bank&quot;, s2 = &quot;kanb&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> For example, swap the first character with the last character of s2 to make &quot;bank&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s1 = &quot;attack&quot;, s2 = &quot;defend&quot;
+<strong>Output:</strong> false
+<strong>Explanation:</strong> It is impossible to make them equal with one string swap.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s1 = &quot;kelb&quot;, s2 = &quot;kelb&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> The two strings are already equal, so no string swap operation is required.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s1.length, s2.length &lt;= 100</code></li>
+	<li><code>s1.length == s2.length</code></li>
+	<li><code>s1</code> and <code>s2</code> consist of only lowercase English letters.</li>
+</ul>

1915-check-if-one-string-swap-can-make-strings-equal/solution.py

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+# Approach 2: Only Check Differences
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def areAlmostEqual(self, s1: str, s2: str) -> bool:
+        first_index_diff = 0
+        second_index_diff = 0
+        num_diffs = 0
+
+        for i in range(len(s1)):
+            if s1[i] != s2[i]:
+                num_diffs += 1
+
+                if num_diffs > 2:
+                    return False
+                elif num_diffs == 1:
+                    first_index_diff = i
+                else:
+                    second_index_diff = i
+
+        return (
+            s1[first_index_diff] == s2[second_index_diff] and
+            s1[second_index_diff] == s2[first_index_diff]
+        )
+        
+