<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:
        """
        - dynamic programming
        - dp[i][j] = shortest common superseq of str1[:i+1] and str2[:j+1]
        -   if str1[i+1]==str2[j+1], then dp[i+1][j+1]=dp[i][j]+str1[i+1]
            else: dp[i+1][j]=dp[i][j]+str1[i+1], and dp[i][j+1]=dp[i][j]+str2[j+1]
            dp[i][j]=dp[i-1][j]+str1[i], and dp[i][j]=dp[i][j-1]+str2[j]
        - base case: dp[0][0]=str1[0]+str2[0] if str1[0]!=str2[0] else str1[0]
        """

        @lru_cache(maxsize=8000)
        def backtrack(i, j):
            if (i==-1) and (j==-1):
                return ""
            elif (i==-1):
                # str1="", str2="a"
                return str2[:j+1]
            elif (j==-1):
                return str1[:i+1]
            if str1[i]==str2[j]:
                return backtrack(i-1, j-1)+str1[i]
            else:
                result1 = backtrack(i-1, j)+str1[i]
                result2 = backtrack(i, j-1)+str2[j]
                if len(result1)<len(result2):
                    return result1
                else:
                    return result2

        return backtrack(len(str1)-1, len(str2)-1)
# TC: O(M*N) where M=len(str1) and N=len(str2)
# SC: O(M*N)
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