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
impl Solution {
    pub fn palindrome_partition(s: String, k: i32) -> i32 {
        let n = s.len();
        let s: Vec<char> = s.chars().collect();
        let mut g = vec![vec![0; n]; n];

        for i in (0..n).rev() {
            for j in i + 1..n {
                g[i][j] = if s[i] != s[j] { 1 } else { 0 };
                if i + 1 < j {
                    g[i][j] += g[i + 1][j - 1];
                }
            }
        }

        let mut f = vec![vec![0; (k + 1) as usize]; n + 1];
        let inf = i32::MAX;

        for i in 1..=n {
            for j in 1..=i.min(k as usize) {
                if j == 1 {
                    f[i][j] = g[0][i - 1];
                } else {
                    f[i][j] = inf;
                    for h in (j - 1)..i {
                        f[i][j] = f[i][j].min(f[h][j - 1] + g[h][i - 1]);
                    }
                }
            }
        }

        f[n][k as usize]
    }
}
