<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) {
        
        function<int(string)>f=[&](string s)
        {
            int p1=0,p2=s.size()-1;
            int ans=0;
            while(p1<p2)
            {
                if(s[p1]!=s[p2]) ans++;
                p1++,p2--;
            }
            return ans;
        };

        int n = s.size();

        vector<vector<int>>dp(n+1,vector<int>(k+1,1e9));
        dp[n][0]=0;

        for(int i=n-1;i>=0;i--)
        {
            for(int x=1;x<=k;x++)
            {
                string str="";
                for(int j=i;j<n;j++)
                {
                    str+=s[j];
                    int cost = f(str);
                    dp[i][x] = min(dp[i][x],cost+dp[j+1][x-1]);
                }
            }
        }
        return dp[0][k];
    }
};
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