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

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

        # populate the dp table to find the lcs subsequence
        for index_1 in reversed(range(n + 1)):
            for index_2 in reversed(range(m + 1)):

                if index_1 == n or index_2 == m:
                    dp[index_1][index_2] = 0
                
                elif str1[index_1] == str2[index_2]:
                    dp[index_1][index_2] = 1 + dp[index_1 + 1][index_2 + 1]
                
                else:
                    dp[index_1][index_2] = max(
                        dp[index_1 + 1][index_2],
                        dp[index_1][index_2 + 1]
                    )

        i, j = 0, 0
        
        lcs_string = ""

        # find the lcs subsequence string
        while i < n and j < m:
            if str1[i] == str2[j]:
                lcs_string += str1[i]
                i += 1
                j += 1
            
            else:
                val_1 = dp[i + 1][j]
                val_2 = dp[i][j + 1]

                if val_1 >= val_2:
                    i += 1

                else:
                    j += 1
        
        ans = ""
        main_ptr_1, main_ptr_2 = 0, 0

        for lcs_ptr in range(len(lcs_string)):

            while str1[main_ptr_1] != lcs_string[lcs_ptr] and main_ptr_1 < n:
                ans += str1[main_ptr_1]
                main_ptr_1 += 1
            
            main_ptr_1 += 1

            while str2[main_ptr_2] != lcs_string[lcs_ptr] and main_ptr_2 < m:
                ans += str2[main_ptr_2]
                main_ptr_2 += 1
        
            main_ptr_2 += 1
            
            ans += lcs_string[lcs_ptr]

        while main_ptr_1 < n:
            ans += str1[main_ptr_1]
            main_ptr_1 += 1

        while main_ptr_2 < m:
            ans += str2[main_ptr_2]
            main_ptr_2 += 1

        return ans
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