<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 collections import deque

class Solution:
    def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
        memo = {}

        l1 = len(str1)
        l2 = len(str2)

        def dp(i, j):
            if (i, j) in memo:
                return memo[(i, j)]

            if i == 0:
                memo[(i, j)] = j
                return j

            if j == 0:
                memo[(i, j)] = i
                return i

            if str1[i - 1] == str2[j - 1]:
                ans = dp(i - 1, j - 1) + 1
            else:
                ans1 = dp(i - 1, j) + 1
                ans2 = dp(i, j - 1) + 1
                ans = min(ans1, ans2)

            memo[(i, j)] = ans
            return ans

        dp(l1, l2)

        ans = deque()

        def traverse(i, j):
            if i == 0:
                ans.extendleft(reversed(str2[:j]))
                return
            if j == 0:
                ans.extendleft(reversed(str1[:i]))
                return

            if str1[i - 1] == str2[j - 1]:
                ans.appendleft(str1[i - 1])
                traverse(i - 1, j - 1)
            elif memo[(i - 1, j)] < memo[(i, j - 1)]:
                ans.appendleft(str1[i - 1])
                traverse(i - 1, j)
            else:
                ans.appendleft(str2[j - 1])
                traverse(i, j - 1)

        traverse(l1, l2)

        return "".join(ans)

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