<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) {
        int s1Len = s1.length();
        int s2Len = s2.length();
        int s3Len = s3.length();

        int i = 0;
        int j = 0;
        int k = 0;
        while (i<s1Len || j<s2Len) {
            if (s1[i] == s3[k]) {
                cout <<"i "<< i << " " << s1[i] << " " << s3[k] << endl;
                i++;
                k++;
            }else if (s2[j] == s3[k]) {
                cout << "j " << j <<" "<< s1[j] << " " << s3[k] << endl;
                j++;
                k++;
            }
            else {
                cout << "here" << endl;
            }
        }
        if (i < s1Len || j < s2Len || k < s3Len) {
            return false;
        }
        return true;
    }
    */
    /*
    bool isInterleave(string s1, string s2, string s3) { //Time limit exceeded
        int s1Len = s1.length();
        int s2Len = s2.length();
        int s3Len = s3.length();

        if(s1Len == 0 && s2Len == 0 && s3Len==0){
              return true;
        }
        if((s1Len == 0 || s2Len == 0) && s3Len==0){
            return false;
        }
        if(s1Len == 0 && s2Len == 0 ){
            return false;
        }
        bool first = false;
        if(s1[0] == s3[0]){
            first = isInterleave(s1.substr(1,s1Len),s2,s3.substr(1,s3Len));
        }
        bool second = false;
        if(s2[0] == s3[0]){
            second = isInterleave(s1,s2.substr(1,s2Len),s3.substr(1,s3Len));
        }
        return (first || second);
    }
    */
    unordered_map<string,bool> inter_map;
    bool isInterleave(string s1, string s2, string s3) { //Time limit exceeded
        int s1Len = s1.length();
        int s2Len = s2.length();
        int s3Len = s3.length();

        if(s1Len == 0 && s2Len == 0 && s3Len==0){
              return true;
        }
        if((s1Len == 0 || s2Len == 0) && s3Len==0){
            return false;
        }
        if(s1Len == 0 && s2Len == 0 ){
            return false;
        }
        if(inter_map.find(s1+','+s2) != inter_map.end()){
            return inter_map[s1+','+s2];
        }
        bool first = false;
        if(s1[0] == s3[0]){
            first = isInterleave(s1.substr(1,s1Len),s2,s3.substr(1,s3Len));
            inter_map[s1.substr(1,s1Len)+','+s2] = first;
        }
        bool second = false;
        if(s2[0] == s3[0]){
            second = isInterleave(s1,s2.substr(1,s2Len),s3.substr(1,s3Len));
            inter_map[s1+','+s2.substr(1,s2Len)] = second;
        }

        return (first || second);
    }
};
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