<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:
        import heapq
        q = [(0, 0, 0, '')]
        N, M = len(str1), len(str2)
        visited = set()
        while q:
            d, i, j, s = heappop(q)
            if i == N and j == M: return s
            elif i == N:
                heappush(q, (d+M-j, i, M, s+str2[j:]))
            elif j == M:
                heappush(q, (d+N-i, N, j, s+str1[i:]))
            elif str1[i] == str2[j]:
                heappush(q, (d+1, i+1, j+1, s+str1[i]))
                visited.add((i+1, j+1))
            else:
                if (i+1, j) not in visited:
                    heappush(q, (d+1, i+1, j, s+str1[i]))
                    visited.add((i+1, j))
                if (i, j+1) not in visited:
                    heappush(q, (d+1, i, j+1, s+str2[j]))
                    visited.add((i, j+1))
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