<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>

from array import array


class Solution:
    def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
        dp = [[array('i') for _ in str2] for _ in str1]
        
        for i, ci in enumerate(str1):
            for j, cj in enumerate(str2):
                dp[i][j] = max(dp[i][j-1] if j > 0 else 0, dp[i-1][j] if i > 0 else 0, (dp[i-1][j-1] if i * j > 0 else 0) + (ci == cj))

        ret, i, j = '', len(str1)-1, len(str2)-1

        while i != -1:
            if j == -1:
                ret += str1[i::-1]
                break
            
            if str1[i] == str2[j]:
                ret += str1[i]
                i, j = i - 1, j - 1
            elif (dp[i][j-1] if j > 0 else 0) == dp[i][j]:
                ret += str2[j]
                j -= 1
            else:
                ret += str1[i]
                i -= 1
        else:
            if j != -1:
                ret += str2[j::-1]
        ret = ret[::-1]

        return ret
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