<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 isInterleave(string s1, string s2, string s3) {
        if (s2.size() < s1.size()) {
        std::string tmp = s1;
        s1 = s2;
        s2 = tmp;
    }
    
    if (s3.size() != s1.size() + s2.size()) return false;
    if(s3.size() == 0) return true;
    
    std::vector<bool> curr(s1.size()+1, false);
    std::vector<bool> prev(s1.size()+1, false);
    prev[0] = true;
    for (int i = 1; i < prev.size(); i++) {
        prev[i] = prev[i-1] && s3[i-1] == s1[i-1];
    }
    
    for (int i = 0; i < s2.size(); i++) {
        curr[0] = prev[0] && s3[i] == s2[i];
        for (int j = 1; j < curr.size(); j++) {
            if (s3[i+j] == s1[j-1] && s3[i+j] == s2[i]) {
                curr[j] = prev[j]|| curr[j-1];
            } else if (s3[i+j] == s2[i]) {
                curr[j] = prev[j];
            } else if (s3[i+j] == s1[j-1]) {
                curr[j] = curr[j-1];
            } else {
                curr[j] = false;
            }
            
            cout << "s1:  "  << s1.substr(0, j) << " s2: " << s2.substr(0, i+1) << " target: " << s3.substr(0, i+j+1) << " " << curr[j] << endl;
            
        }
        prev = curr;
    }
    
    return curr[curr.size()-1];
    }
};
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