<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>

# dp top down
class Solution:
    def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
        m, n = len(str1), len(str2)
        @functools.lru_cache(maxsize=100000) # this will fix the memoery limits
        def dp(i, j): # you want to create str1[i:] and str2[j:] minimum
            # base case
            if i == m or j == n:
                return str1[i:] if j == n else str2[j:] # if both are equal it returns 0, otherwise remainder
            
            # choose one or another if equal choose both
            if str1[i] == str2[j]:
                return str1[i]+dp(i+1, j+1) # use the common chars
            else:
                options1 = str1[i] + dp(i+1, j)
                options2 = str2[j] + dp(i, j+1)
                return options1 if len(options1) <= len(options2) else options2
        
        return dp(0, 0)

        # time: O(N*N * N) for dp
        # space: O(N*N * N) for memoery
            
        
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