<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>

/*
dp[i][k]
[xxxxxxxx][jxxxxxxxi] dp[i][k] = min { dp[j-1][k-1] + cnt(s[j:i]) } for j = k,....,i 
*/
class Solution {
private:
    int cnt(const string str, int startIdx, int endIdx) {
        int curCnt = 0;
        while (endIdx > startIdx) {
            if (str[startIdx] != str[endIdx])
                curCnt++;
            endIdx--;
            startIdx++;
        }
        return curCnt;
    } 
public:
    int palindromePartition(string s, int K) {
        int N = s.size();
        s = "#" + s;

        vector<vector<int>> dp(N + 1, vector<int>(K + 1, INT_MAX / 2));
        // dp[0][0]
        dp[0][0] = 0;

        // dp[0][k]

        // dp[i][0]

        for (int i = 1; i <= N; i++) {
            for (int k = 1; k <= min(i, K); k++) {
                for (int j = i; j >= k; j--) {
                    dp[i][k] = min(dp[i][k], dp[j - 1][k - 1] + cnt(s, j, i));
                }
            }
        }
        return dp[N][K];
    }
};

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