<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, s: str, t: str) -> str:
        @cache
        def helper(i , j):
            if i==len(s) and j==len(t):
                return 0
            
            if i==len(s):
                return len(t)-j
            
            if j==len(t):
                return len(s)-i

            if s[i] == t[j]:
                return helper(i+1 , j+1) + 1
            
            else:
                val1 = helper(i+1 , j) + 1
                val2 = helper(i , j+1) + 1

                return min(val1 , val2)

        def helper2(i , j , cur , result = ''):
            if i==len(s) and j==len(t):
                return result

            if i == len(s):
                return result+t[j:]

            if j==len(t):
                return result+s[i:]

            if s[i] == t[j]:
                return helper2(i+1 , j+1 , cur-1 , result+s[i])
            
            else:
                if helper(i+1 , j) == cur-1:
                    return helper2(i+1 , j , cur-1 , result+s[i])
                else:
                    return helper2(i , j+1 , cur-1 , result+t[j])

        return helper2(0 , 0 , helper(0 , 0) , '')
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