<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:
        dp = [[-1 for _ in range(1001)] for _ in range(1001)]
        
        def rec(i, j):
            if i == len(str1) or j == len(str2):
                return 0
            if dp[i][j] != -1:
                return dp[i][j]
            
            ans = max(rec(i + 1, j), rec(i, j + 1))
            if str1[i] == str2[j]:
                ans = max(ans, 1 + rec(i + 1, j + 1))
            
            dp[i][j] = ans
            return ans
        
        def build_scs(s1, s2):
            ans = []
            i, j = 0, 0
            
            while i < len(s1) and j < len(s2):
                if s1[i] == s2[j]:
                    ans.append(s1[i])
                    i += 1
                    j += 1
                else:
                    if rec(i + 1, j) > rec(i, j + 1):
                        ans.append(s1[i])
                        i += 1
                    else:
                        ans.append(s2[j])
                        j += 1
            
            # Add remaining characters of s1 and s2
            while i < len(s1):
                ans.append(s1[i])
                i += 1
            
            while j < len(s2):
                ans.append(s2[j])
                j += 1
            
            return ''.join(ans)
        
        return build_scs(str1, str2)
        
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