<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:
        def lcsUtil(s1, s2, i, j, dp):
            if i == 0 or j == 0:
                return 0

            if dp[i][j] != -1:
                return dp[i][j]

            if s1[i - 1] == s2[j - 1]:
                dp[i][j] = 1 + lcsUtil(s1, s2, i - 1, j - 1, dp)
            else:
                dp[i][j] = max(lcsUtil(s1, s2, i - 1, j, dp), lcsUtil(s1, s2, i, j - 1, dp))

            return dp[i][j]

        n, m = len(str1), len(str2)
        dp = [[-1 for _ in range(m + 1)] for _ in range(n + 1)]

        # Compute LCS length with memoization
        lcs_length = lcsUtil(str1, str2, n, m, dp)

        # Reconstruct the shortest common supersequence using the DP table
        i, j = n, m
        scs = []

        while i > 0 and j > 0:
            if str1[i - 1] == str2[j - 1]:
                scs.append(str1[i - 1])
                i -= 1
                j -= 1
            elif dp[i - 1][j] > dp[i][j - 1]:
                scs.append(str1[i - 1])
                i -= 1
            else:
                scs.append(str2[j - 1])
                j -= 1

        # Add remaining characters from either string
        while i > 0:
            scs.append(str1[i - 1])
            i -= 1
        while j > 0:
            scs.append(str2[j - 1])
            j -= 1

        scs.reverse()
        return ''.join(scs)

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