<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, text1: str, text2: str) -> str:
        m, n = len(text1), len(text2)
        dp=[[-1]*(1+n) for _ in range(1+m)]

        def memo(i, j, dp):
            if i<0 or j<0: return 0
            if dp[i][j]!= -1: return dp[i][j]
            if text1[i]== text2[j]:
                dp[i][j]= 1+ memo(i-1, j-1, dp)
                return dp[i][j]
            else:
                dp[i][j]= max( memo(i-1, j, dp), memo(i, j-1, dp))
                return dp[i][j]
        
        memo(m-1, n-1, dp)

        string= ""
        i,j=m-1, n-1
        
        while i >= 0 and j >= 0:
            if text1[i] == text2[j]:
                string += text1[i]
                i -= 1
                j -= 1
            elif dp[i][j - 1] > dp[i - 1][j]:
                string += text2[j]
                j -= 1
            else:
                string += text1[i]
                i -= 1

        while i >= 0:
            string += text1[i]
            i -= 1
        while j >= 0:
            string += text2[j]
            j -= 1
            
        return string[::-1]
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