<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:
        dp = {}
        def lcs(i,j):
            if i == -1 or j == -1:
                return 0 
            if (i,j) in dp:
                return dp[(i,j)]
            if str1[i] == str2[j]:
                dp[(i,j)] = 1 + lcs(i-1,j-1)
            else:
                dp[(i,j)] = max(lcs(i,j-1),lcs(i-1,j))
            return dp[(i,j)]
        l = lcs(len(str1) - 1, len(str2) - 1)
        i = len(str1) - 1
        j = len(str2) - 1
        LCS = ""
        indexes_str1 = []
        indexes_str2 = []
        while i >= 0 and j >= 0:
            if str1[i] == str2[j]:
                indexes_str1.append(i)
                indexes_str2.append(j)
                i-=1
                j-=1
            elif i> 0 and dp[(i,j)] == dp[(i-1,j)]:
                i -= 1
            elif j>0 and dp[(i,j)] == dp[(i,j-1)]:
                j-=1
            else:
                break
        indexes_str1.reverse()
        indexes_str2.reverse()
        i = -1
        j = -1
        result = ""
        for index1,index2 in zip(indexes_str1,indexes_str2):
            for x in range(i+1,index1):
                result = result + str1[x]
            for y in range(j+1,index2):
                result = result + str2[y]
            result = result + str1[index1]
            i,j = index1,index2
        for x in range(i+1,len(str1)):
            result = result + str1[x]
        for y in range(j+1,len(str2)):
            result = result + str2[y]
        return result
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