<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:
        has={}
        def lcs(m,n):
            if (m,n) in has:
                return has[(m,n)]
            if m==-1 or n==-1:
                return ""
            if str1[m]==str2[n]:
                return str1[m]+lcs(m-1,n-1)
            first=lcs(m-1,n)
            second=lcs(m,n-1)
            has[(m,n)]=first if len(first)>len(second) else second
            return has[(m,n)]

        l1,l2=len(str1),len(str2)
        common=lcs(l1-1,l2-1)
        common=common[::-1]
        i,j,k=0,0,0
        lk=len(common)
        ans=""
        while i<l1 and j<l2 and k<lk:
            while i<l1 and str1[i]!=common[k]:
                ans+=str1[i]
                i+=1 
            while j<l2 and str2[j]!=common[k]:
                ans+=str2[j]
                j+=1 
            ans+=common[k]
            i+=1 
            j+=1 
            k+=1 
        ans+=str1[i:]+str2[j:]+common[k:]
        return ans
            
        
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