<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:
        n1, n2 = len(str1), len(str2); 
        lenpos = [-1]; lenstr = ['']; stat2 = defaultdict(list)
        for i, c in enumerate(str2): stat2[c].append(i)
        for c in str1:
            if c not in stat2: continue
            for j in stat2[c][::-1]:
                p = bisect.bisect_left(lenpos, j)
                if p == len(lenpos): lenpos.append(j); lenstr.append(lenstr[-1]+c); continue
                if j < lenpos[p]: lenpos[p] = j; lenstr[p] = lenstr[p-1] + c
        res = ''; lcs = lenstr[-1]; i = j = k = 0
        while k < len(lcs):
            while i < n1 and str1[i] != lcs[k]: res += str1[i]; i += 1
            while j < n2 and str2[j] != lcs[k]: res += str2[j]; j += 1
            res += lcs[k]; i += 1; j += 1; k += 1
        res += str1[i:]; res += str2[j:]
        return res
19555