<p>Given two strings <code>str1</code> and <code>str2</code>, return <em>the shortest string that has both </em><code>str1</code><em> and </em><code>str2</code><em> as <strong>subsequences</strong></em>. If there are multiple valid strings, return <strong>any</strong> of them.</p>

<p>A string <code>s</code> is a <strong>subsequence</strong> of string <code>t</code> if deleting some number of characters from <code>t</code> (possibly <code>0</code>) results in the string <code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> str1 = &quot;abac&quot;, str2 = &quot;cab&quot;
<strong>Output:</strong> &quot;cabac&quot;
<strong>Explanation:</strong> 
str1 = &quot;abac&quot; is a subsequence of &quot;cabac&quot; because we can delete the first &quot;c&quot;.
str2 = &quot;cab&quot; is a subsequence of &quot;cabac&quot; because we can delete the last &quot;ac&quot;.
The answer provided is the shortest such string that satisfies these properties.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> str1 = &quot;aaaaaaaa&quot;, str2 = &quot;aaaaaaaa&quot;
<strong>Output:</strong> &quot;aaaaaaaa&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= str1.length, str2.length &lt;= 1000</code></li>
	<li><code>str1</code> and <code>str2</code> consist of lowercase English letters.</li>
</ul>

class Solution:
    def shortestCommonSupersequence(self, s1: str, s2: str) -> str:
        
        m, n = len(s1), len(s2)
        dp = [[()] * (n + 1) for _ in range(m + 1)]

        for i in range(m + 1):
            dp[i][0] = (s1[:i], None, i)
        
        for j in range(n + 1):
            dp[0][j] = (s2[:j], None, j)
        
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if s1[i - 1] == s2[j - 1]:
                    dp[i][j] = (s1[i - 1], (i - 1, j - 1),
                                 dp[i - 1][j - 1][2] + 1)
                else:
                    if dp[i - 1][j][2] < dp[i][j - 1][2]:
                        dp[i][j] = (s1[i - 1], (i - 1, j), 
                                    dp[i - 1][j][2] + 1)
                    else:
                        dp[i][j] = (s2[j - 1], (i, j - 1), 
                                    dp[i][j - 1][2] + 1)

        seq = []
        i, j = m, n
        while True:
            ch, nxt, _ = dp[i][j]
            seq.append(ch)
            if nxt is None:
                break
            i, j = nxt
        
        return ''.join(seq[::-1])

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