<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 = [[(None, 0)] * (len(str2) + 1) for _ in range(len(str1) + 1)]
        for i1 in range(len(str1)):
            dp[i1][len(str2)] = (None, len(str1) - i1, i1)
        for i2 in range(len(str2)):
            dp[len(str1)][i2] = (None, len(str2) - i2, i2)
        
        for i1 in reversed(range(len(str1))):
            for i2 in reversed(range(len(str2))):
                if str1[i1] == str2[i2]:
                    dp[i1][i2] = ((i1 + 1, i2 + 1), dp[i1 + 1][i2 + 1][1], str1[i1])
                    continue
                    
                p1 = ((i1 + 1, i2), dp[i1 + 1][i2][1] + 1, str1[i1])
                p2 = ((i1, i2 + 1), dp[i1][i2 + 1][1] + 1, str2[i2])
                    
                dp[i1][i2] = min(p1, p2, key=lambda x: x[1])
                
        stack = list()
        r = 0
        c = 0
        while dp[r][c][0] is not None:
            stack.append(dp[r][c][2])
            r, c = dp[r][c][0]
            
        if r == len(str1):
            stack.append(str2[c:])
        elif c == len(str2):
            stack.append(str1[r:])
        
        return "".join(stack)
                

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