<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:
        def scs(i1, i2):
            if not i2 in dp[i1]:
                if i1 == len(s1) or i2 == len(s2):
                    dp[i1][i2] = max(len(s1) - i1, len(s2) - i2)
                elif s1[i1] == s2[i2]:
                    dp[i1][i2] = 1 + scs(i1+1, i2+1)
                else:
                    dp[i1][i2] = 1 + min(scs(i1, i2+1), scs(i1+1, i2))

            return dp[i1][i2]

        dp = defaultdict(dict)
        scs(0, 0)
        sc = ''
        i1, i2 = 0, 0

        print(dp)

        while i1 < len(s1) and i2 < len(s2):
            if s1[i1] == s2[i2]:
                sc += s1[i1]
                i1, i2 = i1+1, i2+1
            elif dp[i1+1][i2] < dp[i1][i2+1]:
                sc += s1[i1]
                i1 += 1
            else:
                sc += s2[i2]
                i2 += 1

        if i1 < len(s1):
            sc += s1[i1:]
        elif i2 < len(s2):  
            sc += s2[i2:]

        return sc
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