<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 pal[101][101];
    int dp[101][101];
    int f(string s, int i, int k) {
        if(i >= s.size())
            return ((k == 0) ? 0 : 1011);
        int ans = 1011;
        if(k < 0)
            return ans;
        if(dp[i][k] != -1)
            return dp[i][k];
        for(int l = i; l < s.size(); l++) {
            ans = min(ans, pal[i][l] + f(s, l + 1, k - 1));
        }
        return dp[i][k] = ans;
    }
    
    int dfs(string s, int i, int j) {
        if(i >= j)
            return 0;
        int cnt = 0;
		if(pal[i][j] != -1)
			return pal[i][j];
        if(s[i] != s[j])
            cnt++;
        return pal[i][j] = dfs(s, i + 1, j - 1) + cnt;
    }
    
    int palindromePartition(string s, int k) {
        int n = s.size(); 
        memset(pal, -1, sizeof pal);
        memset(dp, -1, sizeof dp);
        for(int i = 0; i < s.size(); i++) {
            pal[i][i] = 0;
            for(int j = i + 1; j < s.size(); j++) {
                if(pal[i][j] == -1) {
                    pal[i][j] = dfs(s, i, j);
                }
            }
        }
        return f(s, 0, k);
    }
};
99631