<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:
        ln1, ln2 = len(str1), len(str2)
        dp = [['' for _ in range(ln2 + 1)] for _ in range(ln1 + 1)]
        for i in range(ln2 + 1):
            dp[-1][i] = [str2[i:], ln2 - i]
        for i in range(ln1 + 1):
            dp[i][-1] = [str1[i:], ln1 - i]
        dp[-1][-1] = ['', 0]
        for i in range(ln1 - 1, -1, -1):
            for j in range(ln2 - 1, -1, -1):
                if str1[i] == str2[j]:
                    temp = dp[i + 1][j + 1][2] if len(dp[i + 1][j + 1]) == 3 else dp[i + 1][j + 1][1]
                    dp[i][j] = [str1[i], (i + 1, j + 1), 1 + temp]
                    continue
                l1 = dp[i + 1][j][2] if len(dp[i + 1][j]) == 3 else dp[i + 1][j][1]
                l2 = dp[i][j + 1][2] if len(dp[i][j + 1]) == 3 else dp[i][j + 1][1]
                if l1 < l2:
                    dp[i][j] = [str1[i], (i + 1, j), 1 + l1]
                else:
                    dp[i][j] = [str2[j], (i, j + 1), 1 + l2]
        s = ''
        hold = dp[0][0]
        while 1:
            s += hold[0]
            if len(hold) == 2:return s
            x, y = hold[1][0], hold[1][1]
            hold = dp[x][y]
        return s




        
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