<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 dfp(string &s)
    {
        int cnt = 0;
        for(int i=0;i<(s.length()/2);i++)
        {
            if(s[i]!=s[s.length()-1-i])cnt++;
        }
        return cnt;
    }
    int pp3(int index,int k,string s,vector<vector<int>>&dp)
    {
        // base case
        // if i try to further partiton
        if(k==0 and index<s.size()) return 1e8;
        if(k==0)return 0;
        if(index==s.size()-1)
        {
            // since i have to divide the complete string so 
            // if k has become zero or negative thats not quite valid but
            // we have already handled this case above
            if(k>1)return 1e8;
            return 0;
        }
        
        if(dp[index][k]!=-1)return dp[index][k];

        string temp = "";
        int minsteps = 1e8;
        for(int i=index;i<s.size();i++)
        {
            temp = temp + s[i];
            minsteps = min(minsteps,dfp(temp)+pp3(i+1,k-1,s,dp));
        }
        return dp[index][k] = minsteps;
    }
    int palindromePartition(string s, int k) {
        vector<vector<int>>dp(s.size()+1,vector<int>(k+1,-1));
        return pp3(0,k,s,dp);
    }
};
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