<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:
        # solution while tracking strings 

         # memery limit 
        n = len(str1) 
        m = len(str2) 

        dp = [[0] * (m+1) for _ in range(n+1)] 
        res = [[0] * (m+1) for _ in range(n+1)] 


        for r in range(n): 
            for c in range(m): 
                if str1[r] == str2[c]: 
                    dp[r+1][c+1] = dp[r][c] + 1 
                    res[r+1][c+1] = ("d", str1[r])
                else:
                    if dp[r][c+1] > dp[r+1][c]: 
                        dp[r+1][c+1] = dp[r][c+1]
                        res[r+1][c+1] = ("u", str1[r])
                    else: 
                        dp[r+1][c+1] = dp[r+1][c]
                        res[r+1][c+1] = ("l",str2[c]) 
            
        finres = deque([])

        def const(r,c):
            if r == 0: 
                finres.appendleft(str2[:c])
                return 
            elif c == 0: 
                finres.appendleft(str1[:r])
                return 
            dr = res[r][c][0] 
            ch = res[r][c][1] 
            finres.appendleft(res[r][c][1])

            if dr == "d": 
                const(r-1, c-1) 
            elif dr == "u": 
                const(r-1, c) 
            else: 
                const(r,c-1) 
            
        const(n,m) 
        return "".join(finres) 













        # memory limit 
        n = len(str1) 
        m = len(str2) 

        dp = [[0] * (m+1) for _ in range(n+1)] 


        
        for i in range(n): 
            dp[i+1][0] = [0,str1[:i+1]] 

        for i in range(m): 
            dp[0][i+1] = [0, str2[:i+1]]

        dp[0][0] = [0, ""] 

        for r in range(n): 
            for c in range(m): 
                if str1[r] == str2[c]: 
                    dp[r+1][c+1] = [dp[r][c][0] + 1, dp[r][c][1] + str1[r]]
                
                else:

                    if dp[r][c+1][0] > dp[r+1][c][0]: 
                        dp[r+1][c+1] = [dp[r][c+1][0], dp[r][c+1][1] + str1[r]]
                    else: 
                        dp[r+1][c+1] = [dp[r+1][c][0], dp[r+1][c][1] + str2[c]]
        return dp[n][m][1] 
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