<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 {
public:
    bool find(int i,int j,int k,string& s1,string& s2,string& s3,int n1,int n2,int n3,vector<vector<vector<int>>>&dp)
    {
        
        if(i==n1)
        {
            while(j<n2)
            {
                if(s2[j]!=s3[k])
                {
                    return false;
                }
                j++;
                k++;
            }
            return true;
        }

         if(j==n2)
        {
            while(i<n1)
            {
                if(s1[i]!=s3[k])
                {
                    return false;
                }
                i++;
                k++;
            }
            return true;
        }

        if(s1[i]!=s3[k] && s2[j]!=s3[k])
        {
            return false;
        }
    
        if(dp[i][j][k]!=-1) return dp[i][j][k];
        
        if(s1[i]==s3[k] && s3[k]!=s2[j])
        {
            return dp[i][j][k]=find(i+1,j,k+1,s1,s2,s3,n1,n2,n3,dp);
        }
        else if(s2[j]==s3[k] && s1[i]!=s3[k])
        {
            return dp[i][j][k]=find(i,j+1,k+1,s1,s2,s3,n1,n2,n3,dp);
        }
        else 
        {
            return dp[i][j][k]=find(i,j+1,k+1,s1,s2,s3,n1,n2,n3,dp)||find(i+1,j,k+1,s1,s2,s3,n1,n2,n3,dp);
        }
    }
    bool isInterleave(string s1, string s2, string s3) {
        int n1=s1.length(),n2=s2.length(),n3=s3.length();
        if(n1+n2!=n3) return false;
        vector<vector<vector<int>>>dp(n1,vector<vector<int>>(n2,vector<int>(n3,-1)));
        return find(0,0,0,s1,s2,s3,n1,n2,n3,dp);
    }
};
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