<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:
        m = len(str1)
        n = len(str2)

        dp = [[0 for _ in range(n+1)] for _ in range(m+1)]

        str1 = '#' + str1
        str2 = '#' + str2

        for i in range(1, m+1):
            dp[i][0] = i
        for j in range(1, n+1):
            dp[0][j] = j

        for i in range(1, m+1):
            for j in range(1, n+1):
                if str1[i] == str2[j]:
                    dp[i][j] = dp[i-1][j-1] + 1
                else:
                    # dp[i][j] = max(dp[i-1][j-1], dp[i-1][j], dp[i][j-1])
                    # dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + 1
                    dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1

        print(dp)

        result = ''
        # x = len(m), y = len(n)
        # x = m, y = n
        x = m
        y = n
        while x > 0 and y > 0:
            if str1[x] == str2[y]:
                result += str1[x]
                x-=1
                y-=1
            # elif dp[i][j] == dp[i-1][j] + 1:
            elif dp[x][y] == dp[x-1][y] + 1:
                result += str1[x]
                x-=1
            else:
                result += str2[y]
                y-=1

        while x > 0:
            result += str1[x]
            x -= 1
        
        while y > 0:
            result += str2[y]
            y -= 1

        # AttributeError: 'str' object has no attribute 'reverse'
        # return result.reverse()
        # return reverese(result)
        # return reversed(result)
        return result[::-1]
63756