<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:
        #we know the answer will have max_len of len(str1) + len(str2)
        #we get the dp table for longest commonsubstring, construct new string
        #if arrow up we add character from short string, else we add character from the other
        
        s1, s2 = str1, str2
        if len(s1) < len(s2):
            s1, s2 = s2, s1

        dp = [[(0, None) for _ in range(len(s2) + 1)] for _ in range(len(s1) + 1)]
        for i in range(1, len(s1) + 1, 1):
            for j in range(1, len(s2) + 1, 1):
                if s1[i-1] == s2[j-1]:
                    val, _ = dp[i-1][j-1]
                    dp[i][j] = (val + 1, 'd')
                else:
                    val1, _ = dp[i][j-1]
                    val2, _ = dp[i-1][j]
                    if val2 > val1:
                        dp[i][j] = (val2, 'u')
                    else:
                        dp[i][j] = (val1, 'l')
        
        res = ""
        y,x = len(dp) - 1, len(dp[0]) - 1
        while y > 0 and x > 0:
            _, pointer = dp[y][x]
            if pointer == 'l':
                res = s2[x-1] + res
                x -= 1
            elif pointer == 'u':
                res = s1[y-1] + res
                y -= 1
            else:
                res = s1[y-1] + res
                y -=1
                x -=1

        if y == 0 and x != 0:
            res = s2[:x] + res
        if x == 0 and y != 0:
            res = s1[:y] + res
        
        return res
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