Valid Palindrome III
<p>Given a string <code>s</code> and an integer <code>k</code>, return <code>true</code> if <code>s</code> is a <code>k</code><strong>-palindrome</strong>.</p>

<p>A string is <code>k</code><strong>-palindrome</strong> if it can be transformed into a palindrome by removing at most <code>k</code> characters from it.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcdeca&quot;, k = 2
<strong>Output:</strong> true
<strong>Explanation:</strong> Remove &#39;b&#39; and &#39;e&#39; characters.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abbababa&quot;, k = 1
<strong>Output:</strong> true
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code> consists of only lowercase English letters.</li>
	<li><code>1 &lt;= k &lt;= s.length</code></li>
</ul>
String; Dynamic Programming
func isValidPalindrome(s string, k int) bool {
	n := len(s)
	f := make([][]int, n)
	for i := range f {
		f[i] = make([]int, n)
		f[i][i] = 1
	}
	for i := n - 2; i >= 0; i-- {
		for j := i + 1; j < n; j++ {
			if s[i] == s[j] {
				f[i][j] = f[i+1][j-1] + 2
			} else {
				f[i][j] = max(f[i+1][j], f[i][j-1])
			}
			if f[i][j]+k >= n {
				return true
			}
		}
	}
	return false
}