Interleaving String
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0097.Interleaving%20String/images/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>
String; Dynamic Programming
impl Solution {
    #[allow(dead_code)]
    pub fn is_interleave(s1: String, s2: String, s3: String) -> bool {
        let n = s1.len();
        let m = s2.len();

        if s1.len() + s2.len() != s3.len() {
            return false;
        }

        let mut record = vec![vec![-1; m + 1]; n + 1];

        Self::dfs(
            &mut record,
            n,
            m,
            0,
            0,
            &s1.chars().collect(),
            &s2.chars().collect(),
            &s3.chars().collect(),
        )
    }

    #[allow(dead_code)]
    fn dfs(
        record: &mut Vec<Vec<i32>>,
        n: usize,
        m: usize,
        i: usize,
        j: usize,
        s1: &Vec<char>,
        s2: &Vec<char>,
        s3: &Vec<char>,
    ) -> bool {
        if i >= n && j >= m {
            return true;
        }

        if record[i][j] != -1 {
            return record[i][j] == 1;
        }

        // Set the initial value
        record[i][j] = 0;
        let k = i + j;

        // Let's try `s1` first
        if i < n && s1[i] == s3[k] && Self::dfs(record, n, m, i + 1, j, s1, s2, s3) {
            record[i][j] = 1;
        }

        // If the first approach does not succeed, let's then try `s2`
        if record[i][j] == 0
            && j < m
            && s2[j] == s3[k]
            && Self::dfs(record, n, m, i, j + 1, s1, s2, s3)
        {
            record[i][j] = 1;
        }

        record[i][j] == 1
    }
}
