<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, s1: str, s2: str) -> str:
        dp=[[0]*(len(s2)+1) for _ in range(len(s1)+1)]
        for i in range(1,len(s1)+1):
            for j in range(1,len(s2)+1):
                if s1[i-1]==s2[j-1]:
                    dp[i][j]=1+dp[i-1][j-1]
                else:
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1])
        ans=""
        i,j=len(s1),len(s2)
        while(i>0 and j>0):
            if s1[i-1]==s2[j-1]:
                ans+=s1[i-1]
                i-=1
                j-=1
            else:
                if dp[i-1][j]>=dp[i][j-1]:
                    ans+=s1[i-1]
                    i-=1
                else:
                    ans+=s2[j-1]
                    j-=1
        
        while(i):
            ans+=s1[i-1]
            i-=1
        
        while(j):
            ans+=s2[j-1]
            j-=1
        return ans[::-1]
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