Shortest Common Supersequence
<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>
String; Dynamic Programming
func shortestCommonSupersequence(str1 string, str2 string) string {
	m, n := len(str1), len(str2)
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, n+1)
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			if str1[i-1] == str2[j-1] {
				f[i][j] = f[i-1][j-1] + 1
			} else {
				f[i][j] = max(f[i-1][j], f[i][j-1])
			}
		}
	}
	ans := []byte{}
	i, j := m, n
	for i > 0 || j > 0 {
		if i == 0 {
			j--
			ans = append(ans, str2[j])
		} else if j == 0 {
			i--
			ans = append(ans, str1[i])
		} else {
			if f[i][j] == f[i-1][j] {
				i--
				ans = append(ans, str1[i])
			} else if f[i][j] == f[i][j-1] {
				j--
				ans = append(ans, str2[j])
			} else {
				i, j = i-1, j-1
				ans = append(ans, str1[i])
			}
		}
	}
	for i, j = 0, len(ans)-1; i < j; i, j = i+1, j-1 {
		ans[i], ans[j] = ans[j], ans[i]
	}
	return string(ans)
}