<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:
        
        def successors(node):
            i, j = node
            if i < len(str1) and j < len(str2) and str1[i] == str2[j]:
                yield i+1, j+1
                return
            if i < len(str1):
                yield i+1, j
            if j < len(str2):
                yield i, j+1
            
        path = self.bfs(start=(0, 0), goal=(len(str1), len(str2)), successors=successors)
        assert path is not None
        
        result = ''
        for (i0, j0), (i1, j1) in itertools.pairwise(path):
            if i1-i0:
                result += str1[i0]
            else:
                result += str2[j0]
        
        return result
                
    def bfs(self, start, goal, successors):
        came_from = {start: start}
        visited = came_from.keys()
        frontier = [start]
        while frontier:
            if goal in frontier:
                path = [goal]
                while (node := came_from[path[-1]]) != path[-1]:
                    path.append(node)
                return path[::-1]
            new_frontier = []
            for node in frontier:
                for successor in successors(node):
                    if successor not in visited:
                        came_from[successor] = node
                        new_frontier.append(successor)
            frontier = new_frontier
            
         
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