<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 helper(self, str1, str2, i, j, dp, dp1):
        if i == 0 and j == 0:
            dp1[i][j] = (-1, -1, "")
            return 0
        if i == 0 or j == 0:
            if i == 0:
                dp1[i][j] = (-1, -1, str2[0:j])
            else:
                dp1[i][j] = (-1, -1, str1[0:i])
            return i + j
        if dp[i][j] != -1:
            return dp[i][j]

        if str1[i-1] == str2[j-1]:
            dp[i][j] = 1 + self.helper(str1, str2, i-1, j-1, dp, dp1)
            dp1[i][j] = (i-1, j-1, str1[i-1])
        else:
            ele1 = self.helper(str1, str2, i-1, j, dp, dp1)
            ele2 = self.helper(str1, str2, i, j-1, dp, dp1)
            if ele1 <= ele2:
                dp1[i][j] = (i-1, j, str1[i-1])
            else:
                dp1[i][j] = (i, j-1, str2[j-1])
            dp[i][j] = 1 + min(ele1, ele2)

        return dp[i][j]

    def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
        n = len(str1)
        m = len(str2)

        dp = [[-1] * (m + 1) for _ in range(n + 1)]
        dp1 = [[(-1, -1, "") for _ in range(m + 1)] for _ in range(n + 1)]

        self.helper(str1, str2, n, m, dp, dp1)

        # Reconstruct the SCS from dp1
        i, j = n, m
        scs = []
        while i != -1 and j !=-1:
            prev_i, prev_j, char = dp1[i][j]
            if char:
                scs.append(char)
            i, j = prev_i, prev_j

        return ''.join(reversed(scs))
        # return ''.join(scs)

# x

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