<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:
        lcs = self.longestCommonSubSequence(str1, str2)
        res, i, j = "", 0, 0
        for c in lcs: 
            while str1[i] != c:
                res += str1[i]
                i += 1

            while str2[j] != c:
                res += str2[j]
                j += 1

            res += c
            i, j = i + 1, j + 1
        return res + str1[i:] + str2[j:]

    def longestCommonSubSequence(self, str1, str2):
        m, n = len(str1), len(str2)
        dp = [[""] * (n+1) for _ in range(m+1)]
        for i in range(1, m+1):
            for j in range(1, n+1):
                arr = [dp[i-1][j], dp[i][j-1]]
                if str1[i-1] == str2[j-1]:
                    arr.append(dp[i-1][j-1] + str1[i-1]) 
                dp[i][j] = max(arr, key=lambda x: len(x))
        return dp[-1][-1]
29378