<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 {
    typedef long long LL;

public:
    int palindromePartition(string s, int K) {
        int N = s.size();
        // DP table: dp[k][i][j] = minimum number of changes to partition s[i..j] into k palindromic substrings
        vector<vector<vector<LL>>> dp(K + 1, vector<vector<LL>>(N, vector<LL>(N, 1e9)));

        // Base case: 1 palindrome partition
        for (int i = 0; i < N; ++i) {
            dp[1][i][i] = 0;  // Single character is always a palindrome
        }

        // Fill in dp[1] for substrings of length > 1
        for (int i = N - 2; i >= 0; --i) {
            for (int j = i + 1; j < N; ++j) {
                // Number of changes to make s[i..j] a palindrome
                dp[1][i][j] = (s[i] != s[j]) + (i + 1 <= j - 1 ? dp[1][i + 1][j - 1] : 0);
            }
        }

        // Fill in dp[k] for k > 1
        for (int k = 2; k <= K; ++k) {
            for (int i = N - 2; i >= 0; --i) {
                for (int j = i + k - 1; j < N; ++j) {
                    for (int t = i + k - 2; t < j; ++t) {
                        dp[k][i][j] = min(dp[k][i][j], dp[k - 1][i][t] + dp[1][t + 1][j]);
                    }
                }
            }
        }

        // Result: minimum changes for partitioning the whole string into K palindromic substrings
        return dp[K][0][N - 1];
    }
};

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