Maximum Number of Non-overlapping Palindrome Substrings
<p>You are given a string <code>s</code> and a <strong>positive</strong> integer <code>k</code>.</p>

<p>Select a set of <strong>non-overlapping</strong> substrings from the string <code>s</code> that satisfy the following conditions:</p>

<ul>
	<li>The <strong>length</strong> of each substring is <strong>at least</strong> <code>k</code>.</li>
	<li>Each substring is a <strong>palindrome</strong>.</li>
</ul>

<p>Return <em>the <strong>maximum</strong> number of substrings in an optimal selection</em>.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters within a string.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abaccdbbd&quot;, k = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> We can select the substrings underlined in s = &quot;<u><strong>aba</strong></u>cc<u><strong>dbbd</strong></u>&quot;. Both &quot;aba&quot; and &quot;dbbd&quot; are palindromes and have a length of at least k = 3.
It can be shown that we cannot find a selection with more than two valid substrings.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;adbcda&quot;, k = 2
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no palindrome substring of length at least 2 in the string.
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>
Greedy; Two Pointers; String; Dynamic Programming
func maxPalindromes(s string, k int) int {
	n := len(s)
	dp := make([][]bool, n)
	f := make([]int, n)
	for i := 0; i < n; i++ {
		dp[i] = make([]bool, n)
		f[i] = -1
		for j := 0; j < n; j++ {
			dp[i][j] = true
		}
	}
	for i := n - 1; i >= 0; i-- {
		for j := i + 1; j < n; j++ {
			dp[i][j] = s[i] == s[j] && dp[i+1][j-1]
		}
	}
	var dfs func(int) int
	dfs = func(i int) int {
		if i >= n {
			return 0
		}
		if f[i] != -1 {
			return f[i]
		}
		ans := dfs(i + 1)
		for j := i + k - 1; j < n; j++ {
			if dp[i][j] {
				ans = max(ans, 1+dfs(j+1))
			}
		}
		f[i] = ans
		return ans
	}
	return dfs(0)
}