<p>Given a string <code>s</code> and an integer <code>k</code>, partition <code>s</code> into <code>k</code> <strong><span data-keyword="substring-nonempty">substrings</span></strong> such that the letter changes needed to make each substring a <strong>semi-palindrome</strong>&nbsp;are minimized.</p>

<p>Return the <em><strong>minimum</strong> number of letter changes</em> required<em>.</em></p>

<p>A <strong>semi-palindrome</strong> is a special type of string that can be divided into <strong><span data-keyword="palindrome">palindromes</span></strong> based on a repeating pattern. To check if a string is a semi-palindrome:​</p>

<ol>
	<li>Choose a positive divisor <code>d</code> of the string&#39;s length. <code>d</code> can range from <code>1</code> up to, but not including, the string&#39;s length. For a string of length <code>1</code>, it does not have a valid divisor as per this definition, since the only divisor is its length, which is not allowed.</li>
	<li>For a given divisor <code>d</code>, divide the string into groups where each group contains characters from the string that follow a repeating pattern of length <code>d</code>. Specifically, the first group consists of characters at positions <code>1</code>, <code>1 + d</code>, <code>1 + 2d</code>, and so on; the second group includes characters at positions <code>2</code>, <code>2 + d</code>, <code>2 + 2d</code>, etc.</li>
	<li>The string is considered a semi-palindrome if each of these groups forms a palindrome.</li>
</ol>

<p>Consider the string <code>&quot;abcabc&quot;</code>:</p>

<ul>
	<li>The length of <code>&quot;abcabc&quot;</code> is <code>6</code>. Valid divisors are <code>1</code>, <code>2</code>, and <code>3</code>.</li>
	<li>For <code>d = 1</code>: The entire string <code>&quot;abcabc&quot;</code> forms one group. Not a palindrome.</li>
	<li>For <code>d = 2</code>:
	<ul>
		<li>Group 1 (positions <code>1, 3, 5</code>): <code>&quot;acb&quot;</code></li>
		<li>Group 2 (positions <code>2, 4, 6</code>): <code>&quot;bac&quot;</code></li>
		<li>Neither group forms a palindrome.</li>
	</ul>
	</li>
	<li>For <code>d = 3</code>:
	<ul>
		<li>Group 1 (positions <code>1, 4</code>): <code>&quot;aa&quot;</code></li>
		<li>Group 2 (positions <code>2, 5</code>): <code>&quot;bb&quot;</code></li>
		<li>Group 3 (positions <code>3, 6</code>): <code>&quot;cc&quot;</code></li>
		<li>All groups form palindromes. Therefore, <code>&quot;abcabc&quot;</code> is a semi-palindrome.</li>
	</ul>
	</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1: </strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> s = &quot;abcac&quot;, k = 2 </span></p>

<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> 1 </span></p>

<p><strong>Explanation: </strong> Divide <code>s</code> into <code>&quot;ab&quot;</code> and <code>&quot;cac&quot;</code>. <code>&quot;cac&quot;</code> is already semi-palindrome. Change <code>&quot;ab&quot;</code> to <code>&quot;aa&quot;</code>, it becomes semi-palindrome with <code>d = 1</code>.</p>
</div>

<p><strong class="example">Example 2: </strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> s = &quot;abcdef&quot;, k = 2 </span></p>

<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> 2 </span></p>

<p><strong>Explanation: </strong> Divide <code>s</code> into substrings <code>&quot;abc&quot;</code> and <code>&quot;def&quot;</code>. Each&nbsp;needs one change to become semi-palindrome.</p>
</div>

<p><strong class="example">Example 3: </strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> s = &quot;aabbaa&quot;, k = 3 </span></p>

<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> 0 </span></p>

<p><strong>Explanation: </strong> Divide <code>s</code> into substrings <code>&quot;aa&quot;</code>, <code>&quot;bb&quot;</code> and <code>&quot;aa&quot;</code>.&nbsp;All are already semi-palindromes.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= s.length &lt;= 200</code></li>
	<li><code>1 &lt;= k &lt;= s.length / 2</code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
</ul>

class Solution:
    def minimumChanges(self, s: str, k: int) -> int:
        mem = {}
        def dfs(left, right, ops):
            # If the result for the current state is already computed, return it.
            if (left, right, ops) in mem:
                return mem[(left, right, ops)]
            
            ans = 200  # Initialize the answer with a large value.
            
            # If no more partitions are to be made (ops == 0), then compute the changes needed
            # to make the current substring a semi-palindrome.
            if ops == 0:
                n = right - left + 1
                for modulo in range(1, n):
                    if n % modulo == 0:
                        changes = 0
                        start = 0
                        end = (n // modulo) * modulo
                        for res in range(modulo):
                            new_start = start + res
                            if end + res < n:
                                new_end = end + res
                            else:
                                new_end = end - modulo + res
                            while new_start < new_end:
                                if s[left + new_start] != s[left + new_end]:
                                    changes += 1
                                new_start += modulo
                                new_end -= modulo
                        ans = min(ans, changes)
            else:
                # If there are more partitions to be made, then recursively divide the substring
                # and compute the answer.
                for sep in range(left + 1, right - 2 * ops + 1):
                    ans = min(ans, dfs(left, sep, 0) + dfs(sep + 1, right, ops - 1)) # The left substring needs no partition operations, the right substring needs at least partition * 2 characters for the rest partition operations.
            
            # Store the computed result in the memoization dictionary.
            mem[(left, right, ops)] = ans
            return ans

        return dfs(0, len(s) - 1, k - 1)
534