<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 f(string &s, int i,int j)
    {
        int cnt = 0;
        for(; i<j; i++, j--)
        {
            cnt += (s[i] != s[j]);
        }
        return cnt;
    }


    int palindromePartition(string s, int k) {
        int n = s.size();

        int dp[n][k+1]; // dp[i][k] = till i-th character and having k division minimum character change required. 

        for(int i=0;i<n ;i++)
        {
            for(int j = 0; j<k+1;j++)
            {
                dp[i][j] = 1e9;
            }
        }

        if(k == 1)
        {
            return f(s, 0, n-1);
        }

        


        for(int i = 0; i < n; i++)
        {
            dp[i][1] = f(s, 0, i);

            
        }

        for(int K=2; K <= k; K++) // Keep in mind, when we need to consider all possible partition, we have to do like this, defining partition as first loop and then ...
        {
            for(int i = K-1; i < n; i++)  // ending of each partition at i-th character.
            {
                for(int j = K-1; j <= i; j++) // starting of each partition at j-th character... and previous partition k-1 th
                {
                    dp[i][K] = min(dp[i][K], dp[j-1][K-1] + f(s, j, i));
                }
            }
        }


        return dp[n-1][k];

        
    }
};
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