<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, s: str, t: str) -> str:
        n = len(s)
        m = len(t)
        
        # Initialize dp table with 1-based indexing
        dp = [[0] * (m + 1) for _ in range(n + 1)]
        
        # Fill the dp table (Longest Common Subsequence - LCS)
        for i in range(1, n + 1):
            for j in range(1, m + 1):
                if s[i - 1] == t[j - 1]:
                    dp[i][j] = 1 + dp[i - 1][j - 1]
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
        
        # Backtracking to find the shortest common supersequence (SCS)
        def backtrack(x: int, y: int) -> str:
            if x == 0:
                return t[:y]
            if y == 0:
                return s[:x]
            
            if s[x - 1] == t[y - 1]:
                return backtrack(x - 1, y - 1) + s[x - 1]
            elif dp[x - 1][y] >= dp[x][y - 1]:
                return backtrack(x - 1, y) + s[x - 1]
            else:
                return backtrack(x, y - 1) + t[y - 1]
        
        # Start backtracking from the bottom-right corner of dp table
        return backtrack(n, m)

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