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

        dp = [[0] * (m + 1) for _ in range(n + 1)]
        fm = [[0] * (m + 1) for _ in range(n + 1)]
        for i in range(n):
            for j in range(m):
                if str1[i] == str2[j]:
                    dp[i + 1][j + 1] = dp[i][j] + 1
                    fm[i + 1][j + 1] = 1
                else:
                    if dp[i][j + 1] > dp[i + 1][j]:
                        dp[i + 1][j + 1] = dp[i][j + 1]
                        fm[i + 1][j + 1] = 2
                    else:
                        dp[i + 1][j + 1] = dp[i + 1][j]
                        fm[i + 1][j + 1] = 3
        
        st = ""
        i = n
        j = m
        while i > 0 and j > 0:
            if fm[i][j] == 1:
                st += str1[i - 1]
                i -= 1
                j -= 1
            elif fm[i][j] == 2:
                i -= 1
            elif fm[i][j] == 3:
                j -= 1
        
        st = st[::-1]
        ans = ""
        l1 = l2 = l3 = 0
        while l3 < len(st):
            while l1 < len(str1) and str1[l1] != st[l3]:
                ans += str1[l1]
                l1 += 1
            while l2 < len(str2) and str2[l2] != st[l3] :
                ans += str2[l2]
                l2 += 1
            l2 += 1
            ans += str1[l1]
            l1 += 1
            l3 += 1
        while l1 < len(str1):
            ans += str1[l1]
            l1 += 1
        while l2 < len(str2):
            ans += str2[l2]
            l2 += 1
        return ans
            
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