<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:
        n = len(s)

        # 闭区间
        @cache
        def calc_change(left: int, right: int) -> int:
            res = 0
            while left < right:
                if s[left] != s[right]:
                    res += 1
                left += 1
                right -= 1
            return res

        # 子串右端点是 i
        # j 还剩下多少段需要分割
        @cache
        def dfs(i: int, j: int) -> int:
            if j == 0:
                return 0 if i < 0 else inf
            if i < 0:  # 此时 j > 0，还需要分割
                return inf
            res = inf
            # 枚举子串左端点 left
            for left in range(i, -1, -1):
                # 需要计算把 s[left:i+1] 改成回文串的最小修改次数
                change = calc_change(left, i)
                res = min(res, dfs(left-1, j-1) + change)
            return res
        return dfs(n-1, k)
19100