Palindrome Partitioning III
<p>You are given a string <code>s</code> containing lowercase letters and an integer <code>k</code>. You need to :</p>

<ul>
	<li>First, change some characters of <code>s</code> to other lowercase English letters.</li>
	<li>Then divide <code>s</code> into <code>k</code> non-empty disjoint substrings such that each substring is a palindrome.</li>
</ul>

<p>Return <em>the minimal number of characters that you need to change to divide the string</em>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abc&quot;, k = 2
<strong>Output:</strong> 1
<strong>Explanation:</strong>&nbsp;You can split the string into &quot;ab&quot; and &quot;c&quot;, and change 1 character in &quot;ab&quot; to make it palindrome.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aabbc&quot;, k = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong>&nbsp;You can split the string into &quot;aa&quot;, &quot;bb&quot; and &quot;c&quot;, all of them are palindrome.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;leetcode&quot;, k = 8
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= s.length &lt;= 100</code>.</li>
	<li><code>s</code> only contains lowercase English letters.</li>
</ul>
String; Dynamic Programming
func palindromePartition(s string, k int) int {
	n := len(s)
	g := make([][]int, n)
	for i := range g {
		g[i] = make([]int, n)
	}
	for i := n - 1; i >= 0; i-- {
		for j := 1; j < n; j++ {
			if s[i] != s[j] {
				g[i][j] = 1
			}
			if i+1 < j {
				g[i][j] += g[i+1][j-1]
			}
		}
	}
	f := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, k+1)
	}
	for i := 1; i <= n; i++ {
		for j := 1; j <= min(i, k); j++ {
			if j == 1 {
				f[i][j] = g[0][i-1]
			} else {
				f[i][j] = 100000
				for h := j - 1; h < i; h++ {
					f[i][j] = min(f[i][j], f[h][j-1]+g[h][i-1])
				}
			}
		}
	}
	return f[n][k]
}