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

        def sol(ind1, ind2):
            if ind1 < 0 or ind2 < 0:
                return 0
            if dp[ind1][ind2] != -1:
                return dp[ind1][ind2]

            if str1[ind1] == str2[ind2]:
                dp[ind1][ind2] = 1 + sol(ind1-1, ind2-1)
            else:
                dp[ind1][ind2] = max(sol(ind1-1, ind2), sol(ind1, ind2-1))
            return dp[ind1][ind2]

        sol(n-1, m-1)

        i, j = n-1, m-1
        ans = []
        while i >= 0 and j >= 0:
            if str1[i] == str2[j]:
                ans.append(str1[i])
                i -= 1
                j -= 1
            elif i > 0 and (j == 0 or dp[i-1][j] > dp[i][j-1]):
                ans.append(str1[i])
                i -= 1
            else:
                ans.append(str2[j])
                j -= 1

        while i >= 0:
            ans.append(str1[i])
            i -= 1
        while j >= 0:
            ans.append(str2[j])
            j -= 1

        return ''.join(reversed(ans))

53934