<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:
        # dp[i][j]: str1[i:], str2[j:]

        self.dp = dict() # dp[(i, j)]=the shortest length of common super sequence
        self.str1, self.str2 = str1, str2
        self.nextChoice = dict() # nextChoice(i, j) =(whichString, which index)
        self.helper(0, 0)
        res = list()
        i, j = 0, 0 
        while (i, j) in self.nextChoice:
            s, ii, both = self.nextChoice[(i, j)]
            if s == 1:
                s = str1
                i += 1
                if both == 0:
                    j+=1
            else:
                s = str2
                j += 1
                if both == 0:
                    i+=1
            
            res.append(s[ii])
        print(res)
        while i < len(str1):
            res.append(str1[i])
            i+=1
        while j < len(str2):
            res.append(str2[j])
            j+=1

        return "".join(res)
         
    
    def helper(self, i, j):
        if (i, j) in self.dp:
            return self.dp[(i, j)]
        if i >= len(self.str1) and j >= len(self.str2):
            self.dp[(i, j)] = 0
            return self.dp[(i, j)] 
        if i >= len(self.str1): 
            self.dp[(i, j)] = len(self.str2) - j
            return self.dp[(i, j)] 
        if j >= len(self.str2): 
            self.dp[(i, j)] = len(self.str1) - i
            return self.dp[(i, j)] 


        if self.str1[i] == self.str2[j]:
            self.dp[(i, j)] = 1 + self.helper(i+1, j+1)
            self.nextChoice[(i, j)] = (1, i, 0)
            return self.dp[(i, j)]
        else:
            c1 = self.helper(i+1, j)
            c2 = self.helper(i, j+1)
            if c1 <= c2:
                self.dp[(i, j)] = c1 + 1
                self.nextChoice[(i, j)] = (1, i, 1)
            else:
                self.dp[(i, j)] = c2+ 1
                self.nextChoice[(i, j)] = (2, j, 1)
            return self.dp[(i, j)]
           
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