<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 traverse(self, str1, str2, memo):
        for i in range(len(str1)+1):
            for j in range(len(str2)+1):
                if i == 0 or j == 0:
                    memo[i][j] = i + j
                elif str1[i-1] == str2[j-1]:
                    memo[i][j] = 1 + memo[i-1][j-1]
                else:
                    memo[i][j] = 1 + min(memo[i-1][j], memo[i][j-1])
    def construct_str(self, str1, str2, memo):
        i, j = len(str1), len(str2)
        result = []
        
        while i > 0 and j > 0:
            if str1[i-1] == str2[j-1]:
                result.append(str1[i-1])
                i -= 1
                j -= 1
            elif memo[i-1][j] < memo[i][j-1]:
                result.append(str1[i-1])
                i -= 1
            else:
                result.append(str2[j-1])
                j -= 1
        
        # Add remaining characters from both strings
        while i > 0:
            result.append(str1[i-1])
            i -= 1
        while j > 0:
            result.append(str2[j-1])
            j -= 1
        
        return ''.join(result[::-1])

    def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
        memo = [[-1]*(len(str2)+1) for _ in range(len(str1)+1)]
        self.traverse(str1, str2, memo)
        print(memo)
        return self.construct_str(str1, str2, memo)
58845