<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:
        n1, n2 = len(str1), len(str2)
        cache = [[-1 for _ in range(n2 + 1)] for _ in range(n1 + 1)]

        def store(i, j, elem):
            cache[i][j] = elem
            return elem

        def scs(n1: int, n2: int):
            if n1 == 0:
                return store(n1, n2, n2)

            if n2 == 0:
                return store(n1, n2, n1)

            if cache[n1][n2] != -1:
                return cache[n1][n2]

            if str1[n1 - 1] == str2[n2 - 1]:
                return store(n1, n2, 1 + scs(n1 - 1, n2 - 1))

            op1 = 1 + scs(n1 - 1, n2)
            op2 = 1 + scs(n1, n2 - 1)

            return store(n1, n2, min(op1, op2))

        scs(n1, n2)

        finalStr = []
        n1, n2 = len(str1), len(str2)
        while n1 > 0 and n2 > 0:
            if str1[n1 - 1] == str2[n2 - 1]:
                finalStr.append(str1[n1 - 1])
                n1 -= 1
                n2 -= 1
                continue

            if cache[n1 - 1][n2] <= cache[n1][n2 - 1]:
                finalStr.append(str1[n1 - 1])
                n1 -= 1
                continue

            finalStr.append(str2[n2 - 1])
            n2 -= 1
            continue

        while n1 > 0:
            finalStr.append(str1[n1 - 1])
            n1 -= 1

        while n2 > 0:
            finalStr.append(str2[n2 - 1])
            n2 -= 1

        return "".join(finalStr)[::-1]
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