<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) {
        vector<vector<vector<bool>>> dp(s3.length()+1);
        if(s1.length() + s2.length() != s3.length())
            return false;
        for(int i = 0; i <= s3.length(); i++) {
            vector<vector<bool>> v;
            v.resize(s1.length()+1);
            for(int j = 0; j <= s1.length(); j++) {
                for(int k = 0; k <= s2.length(); k++) {
                    v[j].push_back(false);
                }
            }
            dp[i] = (v);
        }
        // for(int i = 0; i < dp.size(); i++) {
        //     for(int j = 0; j < dp[i].size(); j++) {
        //         for(int k = 0; k < dp[i][j].size(); k++) {
        //             cout << dp[i][j][k] << " ";
        //         }
        //         cout << endl;
        //     }
        //     cout << "d"<<endl;
        // }
        // return false;
        for(int i = 0; i <=s1.length(); i++) {
            for(int j = 0; j <=s2.length(); j++)
                dp[0][i][j] = true;
        }
        for(int i = 1; i <= s3.length(); i++) {
            for(int j = 0; j <= s1.length(); j++) {
                for(int k = 0; k <= s2.length(); k++) {
                    // cout << i << " " << j << " " << k << endl;
                    if(j ==0 && k == 0) {
                    // cout << dp[i][j][k] << endl;
                        continue;
                    }
                    if(j == 0) {
                        if(s3[i-1] == s2[k-1])
                            dp[i][j][k] = dp[i-1][j][k-1] | dp[i][j][k];
                    // cout << dp[i][j][k] << endl;
                        continue;
                    }
                    if(k == 0) {
                        if(s3[i-1] == s1[j-1])
                            dp[i][j][k] = dp[i-1][j-1][k] | dp[i][j][k];
                    // cout << dp[i][j][k] << endl;
                        continue;
                    }
                    // cout << dp.size() << " " << dp[i].size() << " " << dp[i][j].size() << endl;
                    if(s3[i-1] == s1[j-1])
                        dp[i][j][k] = dp[i-1][j-1][k] | dp[i][j][k];
                    if(s3[i-1] == s2[k-1])
                        dp[i][j][k] = dp[i-1][j][k-1] | dp[i][j][k];
                    // cout << dp[i][j][k] << endl;
                }
            }
        }
        bool ans = dp[s3.length()][s1.length()][s2.length()];
        return ans;
    }
};
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