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
function palindromePartition(s: string, k: number): number {
    const n = s.length;
    const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
    for (let i = n - 1; i >= 0; i--) {
        for (let j = i + 1; j < n; j++) {
            g[i][j] = s[i] !== s[j] ? 1 : 0;
            if (i + 1 < j) {
                g[i][j] += g[i + 1][j - 1];
            }
        }
    }
    const f: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
    for (let i = 1; i <= n; i++) {
        for (let j = 1; j <= Math.min(i, k); j++) {
            if (j === 1) {
                f[i][j] = g[0][i - 1];
            } else {
                f[i][j] = 1 << 30;
                for (let h = j - 1; h < i; h++) {
                    f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
                }
            }
        }
    }
    return f[n][k];
}
