<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:
        if len(str1)>=len(str2):
            s1,s2=str1,str2
        else:
            s1,s2=str2,str1
        @cache
        def dfs(i,j,s1,s2):
            if j==-1:
                return 0
            if i==-1:
                return j+1
            if s1[i]==s2[j]:
                return dfs(i-1,j-1,s1,s2)
            return min(dfs(i-1,j,s1,s2),dfs(i,j-1,s1,s2)+1)
        res = []
        p1,p2=len(s1)-1,len(s2)-1
        while p1>=0 and p2>=0:
            if s1[p1]==s2[p2]:
                res.append(s1[p1])
                p1-=1
                p2-=1
                continue
            if dfs(p1-1,p2,s1,s2)<=dfs(p1,p2-1,s1,s2)+1:
                res.append(s1[p1])
                p1-=1
            else:
                res.append(s2[p2])
                p2-=1
        while p1>=0:
            res.append(s1[p1])
            p1-=1
        while p2>=0:
            res.append(s2[p2])
            p2-=1
        res.reverse()


        return ''.join(res)

            

        
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