<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:
        q = collections.deque([(0,0,"")]) # Tuple(str1_len, str2_len, char)
        visited = set([(0,0)])
        parent = {(0,0,""): None}
        w1, w2 = len(str1), len(str2)
        while q:
            i,j,c = q.popleft()
            if i == w1 and j == w2:
                return process((i,j,c), parent)
            
            next_steps = []
            c1 = str1[i] if i<w1 else ""
            c2 = str2[j] if j<w2 else ""

            if c1 == c2:
                next_steps.append((i+1, j+1, c1))
            else:
                next_steps.append((i+1, j, c1))
                next_steps.append((i, j+1, c2))
            
            for i2,j2,c2 in next_steps:
                if (i2,j2) in visited:
                    continue
                if i2 > w1 or j2 > w2:
                    continue                
                q.append((i2,j2,c2))
                visited.add((i2,j2))
                parent[(i2,j2,c2)] = (i,j,c)
        
        raise Exception("Should never come here")


def process(t, parent):
    result = []
    while t is not None:
        _,_,c = t
        result.append(c)
        t = parent.get(t)
    result.reverse()
    return "".join(result)
        
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