<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:
        n1 = len(str1)
        n2 = len(str2)
        dp = [[0]*(n2+1) for _ in range(n1+1)]

        for i in reversed(range(n1)):
            for j in reversed(range(n2)):
                dp[i][j] = max(dp[i+1][j],dp[i][j+1])
                if str1[i] == str2[j]:
                    dp[i][j] = max(dp[i][j],dp[i+1][j+1]+1)

        lcs = dp[0][0]

        i = 0
        j = 0
        ans = ""
        nt1 = [1]*(n1)
        nt2 = [1]*(n2)
        while(i < n1 and j < n2):
            nxt = i,j
            if str1[i] == str2[j]:
                if dp[i][j] == 1 + dp[i+1][j+1]:
                    nxt = i+1,j+1
                    nt1[i] = 0
                    nt2[j] = 0
                    ans += str1[i]
            else:
                if(dp[i+1][j] > dp[i][j+1]):
                    ans += str1[i]
                    nxt = i+1,j
                else:
                    ans += str2[j]
                    nxt = i,j+1
            
            i,j = nxt
        
        ans += str1[i:]
        ans += str2[j:]

        return ans
24466