<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>

class Solution:

    #s1          d    b      b     c     a s2
    # False False False False False False 
    #a False False False False False False 
    #a False False False False False False 
    #b False False False False False False 
    #c False False False False False False 
    #c False False False False False True

    # def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        
    #     if len(s3) != len(s1) + len(s2):
    #         return False 
        
    #     dp = [[False] * (len(s2) + 1) for i in range(len(s1) + 1)]
    #     dp[len(s1)][len(s2)] = True

    #     for i in range(len(s1), -1, -1):
    #         for j in range(len(s2), -1, -1):
    #             if i < len(s1) and s3[i + j] == s1[i] and dp[i + 1][j]:
    #                 dp[i][j] = True 
    #             if j < len(s2) and s3[i + j] == s2[j] and dp[i][j + 1]:
    #                 dp[i][j] = True
    #     return dp[0][0]
        


        

    # Neetcode 150 9/16/2024
    # Time O(s1 * s2) Space O(s1 * s2)
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        
        cache = {} # (idx1, idx2)
        if len(s1) + len(s2) != len(s3):
            return False 

        def dfs(idx1, idx2):
            if idx1 >= len(s1) and idx2 >= len(s2):
                return True 
            if idx1 >= len(s1):
                return s2[idx2:] == s3[idx1 + idx2:]
            if idx2 >= len(s2):
                return s1[idx1:] == s3[idx1 + idx2:]
            if (idx1, idx2) in cache:
                return cache[(idx1, idx2)]

            res = False 
            if s1[idx1] == s3[idx1 + idx2] and s2[idx2] == s3[idx1 + idx2]:
                res1 = dfs(idx1 + 1, idx2)
                res2 = dfs(idx1, idx2 + 1)
                res = res1 or res2
            elif s1[idx1] == s3[idx1 + idx2]:
                res = dfs(idx1 + 1, idx2)
            elif s2[idx2] == s3[idx1 + idx2]:
                res = dfs(idx1, idx2 + 1)
            cache[(idx1, idx2)] = res
            return cache[(idx1, idx2)]

    
        return dfs(0, 0)
18200