Minimum Changes to Make K Semi-palindromes
<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>
Two Pointers; String; Dynamic Programming
class Solution {
public:
    int minimumChanges(string s, int k) {
        int n = s.size();
        int g[n + 1][n + 1];
        int f[n + 1][k + 1];
        memset(g, 0x3f, sizeof(g));
        memset(f, 0x3f, sizeof(f));
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = i; j <= n; ++j) {
                int m = j - i + 1;
                for (int d = 1; d < m; ++d) {
                    if (m % d == 0) {
                        int cnt = 0;
                        for (int l = 0; l < m; ++l) {
                            int r = (m / d - 1 - l / d) * d + l % d;
                            if (l >= r) {
                                break;
                            }
                            if (s[i - 1 + l] != s[i - 1 + r]) {
                                ++cnt;
                            }
                        }
                        g[i][j] = min(g[i][j], cnt);
                    }
                }
            }
        }
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                for (int h = 0; h < i - 1; ++h) {
                    f[i][j] = min(f[i][j], f[h][j - 1] + g[h + 1][i]);
                }
            }
        }
        return f[n][k];
    }
};