<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:
        stack, l1, l2 = [(-1, "")], len(str1), len(str2)
        for i, c in enumerate(str2):
            ptr, start = stack[-1][0]+1, len(stack)-1
            while ptr < l1 and str1[ptr] != c:
                ptr += 1
            if ptr < l1:
                stack.append((ptr, stack[-1][1]+c))
            for j in range(start, 0, -1):
                ptr = stack[j-1][0]+1
                while ptr < stack[j][0] and str1[ptr] != c:
                    ptr += 1
                if ptr < stack[j][0]:
                    stack[j] = (ptr, stack[j-1][1]+c)
        ptr1, ptr2, res = 0, 0, ""
        for c in stack[-1][1]:
            while str1[ptr1] != c:
                res += str1[ptr1]
                ptr1 += 1
            while str2[ptr2] != c:
                res += str2[ptr2]
                ptr2 += 1
            res += c
            ptr1 += 1
            ptr2 += 1
        while ptr1 < l1:
            res += str1[ptr1]
            ptr1 += 1
        while ptr2 < l2:
            res += str2[ptr2]
            ptr2 += 1
        return res
19555