<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:
        l1, l2 = len(str1), len(str2)

        def lcs():
            dp = {}

            for i in range(l1 - 1, -1, -1):
                for j in range(l2 - 1, -1, -1):
                    if str1[i] == str2[j]:
                        dp[(i, j)] = 1 + (dp[(i + 1, j + 1)] if i + 1 < l1 and j + 1 < l2 else 0)
                    else:
                        dp[(i, j)] = max((dp[(i + 1, j)] if i + 1 < l1 else 0), (dp[(i, j + 1)] if j + 1 < l2 else 0))

            return dp

        dp = lcs()

        i, j = 0, 0
        res = ""
        while i < l1 and j < l2:
            if str1[i] == str2[j]:
                res += str1[i]
                i += 1
                j += 1
            else:
                if dp.get((i, j + 1), 0) > dp.get((i + 1, j), 0):
                    res += str2[j]
                    j += 1
                else:
                    res += str1[i]
                    i += 1

        while i < l1:
            res += str1[i]
            i += 1

        while j < l2:
            res += str2[j]
            j += 1

        return res

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