<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:
    def palindromePartition(self, s: str, k: int) -> int:
        def minChanges(i,j):
            changes = 0
            while i < j:
                if s[i] != s[j]:
                    changes +=1 
                i += 1
                j -= 1
            return changes 
        
        changes = {}
        for i in range(len(s)):
            for j in range(i,len(s)):
                changes[(i,j)] = minChanges(i,j)
        dp = {}
        def rec(i,k):
            if i == len(s) and k == 0:
                return 0
            if k < 0 or i >= len(s):
                return float('inf')
            if (i,k) in dp:
                return dp[(i,k)]
            result = float('inf')
            for z in range(i,len(s)):
                ch = changes[(i,z)]
                result = min(result,ch + rec(z+1,k-1))
            dp[(i,k)] = result 
            return result 
        return rec(0,k)
19200