<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, str1: str, str2: str) -> str:
        # Initialize memoization table with -1
        memo = [[-1 for _ in range(len(str2) + 1)] for _ in range(len(str1) + 1)]
        
        def dfs(i: int, j: int) -> int:
            if i == len(str1) and j == len(str2):
                return 0
            if i == len(str1):
                return len(str2) - j
            if j == len(str2):
                return len(str1) - i
            if memo[i][j] != -1:
                return memo[i][j]
            if str1[i] == str2[j]:
                memo[i][j] = dfs(i + 1, j + 1)
            else:
                take1 = dfs(i + 1, j)
                take2 = dfs(i, j + 1)
                memo[i][j] = 1 + min(take1, take2)
            return memo[i][j]
        
        def helper(i: int, j: int, ans: list) -> None:
            if i == len(str1) and j == len(str2):
                return
            if i == len(str1):
                ans.append(str2[j:])
                return
            if j == len(str2):
                ans.append(str1[i:])
                return
            if str1[i] == str2[j]:
                ans.append(str1[i])
                helper(i + 1, j + 1, ans)
            else:
                take1 = dfs(i + 1, j)
                take2 = dfs(i, j + 1)
                if take1 <= take2:
                    ans.append(str1[i])
                    helper(i + 1, j, ans)
                else:
                    ans.append(str2[j])
                    helper(i, j + 1, ans)

        dfs(0, 0)
        ans = []
        helper(0, 0, ans)
        return ''.join(ans)

68668