<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 = [[-1 for i in range(1005)] for i in range(1005)]
        
        def longest(x,y):
    
            if(x<0 or y<0):
                return 0
            if(dp[x][y]!=-1):
                return dp[x][y]
            if(str1[x]==str2[y]):
                dp[x][y]=longest(x-1,y-1)+1
            else:
                dp[x][y]=max(longest(x-1,y),longest(x,y-1))
        
            return dp[x][y]
        
    
        x=len(str1)-1
        y=len(str2)-1
        ans=[]
        
        while(x>=0 and y>=0):
            if(str1[x]==str2[y]):
                ans.append(str1[x])
                x-=1
                y-=1
            elif(longest(x-1,y)>longest(x,y-1)):
                x-=1
            else:
                y-=1
        ans.reverse()
       
        x=0
        y=0
        ptr=0
        dns=[]
        print(ans)
        while(x<len(str1) and y<len(str2) and ptr<len(ans)):
            if(str1[x]!=ans[ptr]):
                dns.append(str1[x])
                x+=1
                continue
            if(str2[y]!=ans[ptr]):
                dns.append(str2[y])
                y+=1
                continue
            dns.append(str1[x])
            x+=1
            y+=1
            ptr+=1
        while(x<len(str1)):
            dns.append(str1[x])
            x+=1
        while(y<len(str2)):
            dns.append(str2[y])
            y+=1
        return ''.join(map(str,dns))
                
        


        
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