<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, s1: str, s2: str) -> str:
        n1, n2 = len(s1), len(s2)

        @lru_cache(maxsize=None)
        def get_lcs(i1, i2):
            if i1 >= n1 or i2 >= n2:
                return 0, ""
            if s1[i1] == s2[i2]:
                lcs_len, subseq = get_lcs(i1+1, i2+1)
                return 1 + lcs_len, s1[i1] + subseq
            else:
                skip_i1, subeq_by_skipping_i1 = get_lcs(i1, i2+1)
                skip_i2, subeq_by_skipping_i2 = get_lcs(i1+1, i2)
                if skip_i1 > skip_i2:
                    return skip_i1, subeq_by_skipping_i1
                else:
                    return skip_i2, subeq_by_skipping_i2

        lcs_len, lcs = get_lcs(0, 0)
        # Now construct the shortest common supersequence
        i, j = 0, 0
        scs = []

        for c in lcs:
            while s1[i] != c:
                scs.append(s1[i])
                i += 1
            while s2[j] != c:
                scs.append(s2[j])
                j += 1
            scs.append(c)
            i += 1
            j += 1

        scs.append(s1[i:])
        scs.append(s2[j:])

        return ''.join(scs)
                
            

            

        
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