<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:
        """
        DP
        time: O(n * m)
        space: O(n * m)
        """
        # Find the Longest Common Subsequence - LCS
        n, m = len(str1), len(str2)
        dp = [[""] * (m + 1) for _ in range(n + 1)] # dp[i][j]: LCS till i,j
        for i in range(1, n + 1):
            for j in range(1, m + 1):
                if str1[i - 1] == str2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + str1[i - 1]
                else:
                    if len(dp[i - 1][j]) > len(dp[i][j - 1]):
                        dp[i][j] = dp[i - 1][j]
                    else:
                        dp[i][j] = dp[i][j - 1]
        # build the SCS
        i, j = n, m
        lcs = dp[n][m]
        result = []
        p, q = 0, 0

        for c in lcs:
            # Add char from str1 until match
            while p < n and str1[p] != c:
                result.append(str1[p])
                p += 1
            # Add char from str2 until match
            while q < m and str2[q] != c:
                result.append(str2[q])
                q += 1
            result.append(c)
            p += 1
            q += 1
        # Add remain char
        result.append(str1[p:])
        result.append(str2[q:])
        return "".join(result)
        
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