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

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if str1[i - 1] == str2[j - 1]:
                    lcs_length[i][j] = 1 + lcs_length[i - 1][j - 1]
                elif lcs_length[i - 1][j] > lcs_length[i][j - 1]:
                    lcs_length[i][j] = lcs_length[i - 1][j]
                else:
                    lcs_length[i][j] = lcs_length[i][j - 1]

        lcs = []
        def construct_lcs(i: int, j: int):
            if i == 0 or j == 0:
                return

            if str1[i - 1] == str2[j - 1]:
                construct_lcs(i - 1, j - 1)
                lcs.append(str1[i - 1])
            elif lcs_length[i][j] == lcs_length[i - 1][j]:
                construct_lcs(i - 1, j)
            else:
                construct_lcs(i, j - 1)

        construct_lcs(m, n)
        lcs = "".join(lcs)

        # scs is obtained by inserting the non-lcs chars into the lcs
        # while preserving their original order
        scs = ""
        str1_pointer = 0
        str2_pointer = 0

        def add_characters(char: str):
            nonlocal str1_pointer, str2_pointer, scs

            while True:
                is_reached = True

                if str1_pointer < m and str1[str1_pointer] != char:
                    scs += str1[str1_pointer]
                    str1_pointer += 1
                    is_reached = False

                if is_reached:
                    str1_pointer += 1
                    break

            while True:
                is_reached = True

                if str2_pointer < n and str2[str2_pointer] != char:
                    scs += str2[str2_pointer]
                    str2_pointer += 1
                    is_reached = False

                if is_reached:
                    str2_pointer += 1
                    break

            scs += char

        for i in range(lcs_length[m][n]):
            char = lcs[i]
            add_characters(char)

        while str1_pointer < m:
            scs += str1[str1_pointer]
            str1_pointer += 1

        while str2_pointer < n:
            scs += str2[str2_pointer]
            str2_pointer += 1

        return scs

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