<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>

class Solution {
public:
    int palindromePartition(string s, int k) {
        const int kInf = s.length();
        vector<vector<vector<int>>> dp(s.length(), vector<vector<int>>(s.length(), vector<int>(k + 1, kInf)));

        auto make_palindrome = [&](int start, int end) -> int {
            int count = 0;
            while (start < end) {
                if (s[start] != s[end]) {
                    count++;
                }
                start++;
                end--;
            }
            return count;
        };

        for (int i = 0; i < s.length(); i++) {
            for (int j = i; j < s.length(); j++) {
                dp[i][j][1] = make_palindrome(i, j);
            }
        }

        for (int partition = 2; partition <= k; partition++) {
            for (int i = 0; i < s.length(); i++) {
                for (int j = i; j < s.length(); j++) {
                    for (int s = i; s < j; s++) {
                        dp[i][j][partition] = min(dp[i][j][partition], dp[i][s][1] + dp[s + 1][j][partition - 1]);
                    }
                }
            }
        }
        return dp[0][s.length() - 1][k];
    }
};
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