<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:
        
        s = str1
        t = str2
        n = len(s)
        m = len(t)
        dp = [[-1]*(m+1) for _ in range(n+1)]
        dp[-1] = [0]*(m+1)
        for i in range(n+1):
            dp[i][-1] = 0
        for i in range(n-1,-1,-1):
            for j in range(m-1,-1,-1):
                if s[i]==t[j]:
                    dp[i][j] = 1+dp[i+1][j+1]
                else:
                    dp[i][j] = max(dp[i+1][j],dp[i][j+1]) 
        ans = ''
        i,j = 0,0
        while(i<n and j<m):
            if s[i]==t[j]:
                ans += s[i]
                i += 1
                j += 1
            else:
                a = 0
                b = 0
                if i+1<n:
                    a = dp[i+1][j]
                if j+1<m:
                    b = dp[i][j+1]
                if a>=b:
                    ans += s[i]
                    i += 1
                else:
                    ans += t[j]   
                    j += 1
        print(ans)
        print(i,j)            
        if i<n:
            ans += s[i:]
        if j<m:
            ans += t[j:]    
        print(dp)                                         
        return ans
        
53934