<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>

def lcs(s1, s2):
    mtx = []
    l1 = len(s1)
    l2 = len(s2)
    for i1 in range(len(s1) + 1):
        mtx.append([None] * (len(s2)+1))
    mtx[0][0] = [0,"",None]
    for i in range(l1):
        mtx[i+1][0] = [i+1,s1[i],(i,0)]
    for i in range(l2):
        mtx[0][i+1] = [i+1,s2[i],(0,i)]
    for i in range(l1):
        for j in range(l2):
            if s1[i] == s2[j]:
                mtx[i+1][j+1] = [mtx[i][j][0]+1, s1[i], (i,j)]
            elif mtx[i][j+1][0] < mtx[i+1][j][0]:
                mtx[i+1][j+1] = [mtx[i][j+1][0]+1, s1[i], (i,j+1)]
            else:
                mtx[i+1][j+1] = [mtx[i+1][j][0]+1, s2[j], (i+1,j)]
    i,j = l1,l2
    rep = ""
    while True:
        rep += mtx[i][j][1]
        ij = mtx[i][j][2]
        if not ij:
            return rep[::-1]
        i,j = ij

class Solution:
    def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
        return lcs(str1, str2)
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