<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>

class Solution:
    def maxPalindromes(self, s: str, k: int) -> int:
        n = len(s)
        
        # Step 1: Create the palindrome DP table
        is_palindrome = [[False] * n for _ in range(n)]
        
        # All single characters are palindromes
        for i in range(n):
            is_palindrome[i][i] = True
        
        # Check for two-character palindromes
        for i in range(n - 1):
            if s[i] == s[i + 1]:
                is_palindrome[i][i + 1] = True
        
        # Fill the DP table for palindromes of length > 2
        for length in range(3, n + 1):
            for i in range(n - length + 1):
                j = i + length - 1
                if s[i] == s[j] and is_palindrome[i + 1][j - 1]:
                    is_palindrome[i][j] = True
        
        # Step 2: Create the DP array for max palindromes
        dp = [0] * n
        
        for i in range(n):
            # Option 1: Do not take any new palindrome ending at i
            if i > 0:
                dp[i] = dp[i - 1]
            
            # Option 2: Check all possible substrings ending at i and having length >= k
            for j in range(i - k + 1, -1, -1):
                if is_palindrome[j][i]:
                    # If this is a valid palindrome, update dp[i]
                    if j == 0:
                        dp[i] = max(dp[i], 1)  # The whole substring is a palindrome
                    else:
                        dp[i] = max(dp[i], dp[j - 1] + 1)
        
        return dp[n - 1]
48000