<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:
        @cache
        def f(i,j):
            if i == len(str1) and j == len(str2):
                return 0,i,j
            if i == len(str1):
                l,*_ = f(i,j+1)
                return l+1,i,j+1
            if j == len(str2):
                l,*_ = f(i+1,j)
                return l+1,i+1,j
            if str1[i] == str2[j]:
                l,*_ = f(i+1,j+1)
                return l+1,i+1,j+1
            li,*_ = f(i+1,j)
            lj,*_ = f(i,j+1)
            return (li+1,i+1,j) if li < lj else (lj+1,i,j+1)
        i = j = 0
        s = []
        while i < len(str1) or j < len(str2):
            l,i1,j1 = f(i,j)
            s.append(str1[i] if i1 != i else str2[j])
            i,j = i1,j1
        return ''.join(s)
240561