[LeetCode Sync] Runtime - 45 ms (55.31%), Memory - 18.5 MB (22.87%)

Author: NinePiece2

Solutions/0097-interleaving-string/README.md

@@ -0,0 +1,51 @@
+<p>Given strings <code>s1</code>, <code>s2</code>, and <code>s3</code>, find whether <code>s3</code> is formed by an <strong>interleaving</strong> of <code>s1</code> and <code>s2</code>.</p>
+
+<p>An <strong>interleaving</strong> of two strings <code>s</code> and <code>t</code> is a configuration where <code>s</code> and <code>t</code> are divided into <code>n</code> and <code>m</code> <span data-keyword="substring-nonempty">substrings</span> respectively, such that:</p>
+
+<ul>
+	<li><code>s = s<sub>1</sub> + s<sub>2</sub> + ... + s<sub>n</sub></code></li>
+	<li><code>t = t<sub>1</sub> + t<sub>2</sub> + ... + t<sub>m</sub></code></li>
+	<li><code>|n - m| &lt;= 1</code></li>
+	<li>The <strong>interleaving</strong> is <code>s<sub>1</sub> + t<sub>1</sub> + s<sub>2</sub> + t<sub>2</sub> + s<sub>3</sub> + t<sub>3</sub> + ...</code> or <code>t<sub>1</sub> + s<sub>1</sub> + t<sub>2</sub> + s<sub>2</sub> + t<sub>3</sub> + s<sub>3</sub> + ...</code></li>
+</ul>
+
+<p><strong>Note:</strong> <code>a + b</code> is the concatenation of strings <code>a</code> and <code>b</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/09/02/interleave.jpg" style="width: 561px; height: 203px;" />
+<pre>
+<strong>Input:</strong> s1 = &quot;aabcc&quot;, s2 = &quot;dbbca&quot;, s3 = &quot;aadbbcbcac&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> One way to obtain s3 is:
+Split s1 into s1 = &quot;aa&quot; + &quot;bc&quot; + &quot;c&quot;, and s2 into s2 = &quot;dbbc&quot; + &quot;a&quot;.
+Interleaving the two splits, we get &quot;aa&quot; + &quot;dbbc&quot; + &quot;bc&quot; + &quot;a&quot; + &quot;c&quot; = &quot;aadbbcbcac&quot;.
+Since s3 can be obtained by interleaving s1 and s2, we return true.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s1 = &quot;aabcc&quot;, s2 = &quot;dbbca&quot;, s3 = &quot;aadbbbaccc&quot;
+<strong>Output:</strong> false
+<strong>Explanation:</strong> Notice how it is impossible to interleave s2 with any other string to obtain s3.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s1 = &quot;&quot;, s2 = &quot;&quot;, s3 = &quot;&quot;
+<strong>Output:</strong> true
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>0 &lt;= s1.length, s2.length &lt;= 100</code></li>
+	<li><code>0 &lt;= s3.length &lt;= 200</code></li>
+	<li><code>s1</code>, <code>s2</code>, and <code>s3</code> consist of lowercase English letters.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow up:</strong> Could you solve it using only <code>O(s2.length)</code> additional memory space?</p>

Solutions/0097-interleaving-string/solution.py

@@ -0,0 +1,19 @@
+class Solution:
+    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
+        m, n = len(s1), len(s2)
+        if m + n != len(s3):
+            return False
+
+        @cache
+        def dfs(i: int, j: int) -> bool:
+            if i >= m and j >= n:
+                return True
+            
+            k = + i + j
+            if i < m and s1[i] == s3[k] and dfs(i + 1, j):
+                return True
+            if j < n and s2[j] == s3[k] and dfs(i, j + 1):
+                return True
+            return False
+        
+        return dfs(0, 0)