<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 count(string s,int i,int j, vector<vector<int>>& dp1)
{
    if(i>=j)
    return 0;
    else if(dp1[i][j]!=-1)
    return dp1[i][j];
    else if(s[i]==s[j])
    return dp1[i][j]=count(s,i+1,j-1,dp1);
    else
    return dp1[i][j]=count(s,i+1,j-1,dp1)+1;
}
int fun(int i,string s,int n,int k,int no,vector<vector<int>>& dp, vector<vector<int>>& dp1)
{
     if(no==k-1)
    {
        return dp[i][no]=count(s,i,s.size()-1,dp1);
    }
    else if(i>=n)
    {
      return 0;
    }
    else if(dp[i][no]!=-1)
    return dp[i][no];
    else {
        int l=INT_MAX;
        int ans=INT_MAX;
        for(int j=i;j<(n+no-k+1);j++)
        {
            dp[j+1][no+1]=fun(j+1,s,n,k,no+1,dp,dp1);
            l=count(s,i,j,dp1)+dp[j+1][no+1];
            ans=min(ans,l);
        }
        return dp[i][no]=ans;
    }
}
    int palindromePartition(string s, int k) {
        int n=s.size();
        vector<vector<int>> dp(101,vector<int>(101,-1));
        vector<vector<int>> dp1(n,vector<int>(n,-1));
        return fun(0,s,n,k,0,dp,dp1);
    }
};
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