<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)
        dp = [ [False] * n for _ in range(n)]

        for i in range(n):
            dp[i][i] = True
        
        for i in range(n-1):
            if s[i] == s[i+1]:
                dp[i][i+1] = True
        
        for length in range(3, n+1):
            for i in range(n-length+1):
                if s[i] == s[i+length-1] and dp[i+1][i+length-2] == True:
                    dp[i][i+length-1] = True
        dp_res = [0] * (n+1)
        for i in range(1, n+1):
            dp_res[i] = dp_res[i-1]  # If we don't take any palindrome ending at position i-1
            for j in range(i-k, -1, -1):  # Look for palindromes of length >= k
                if dp[j][i-1]:
                    dp_res[i] = max(dp_res[i], dp_res[j] + 1)
                    break  # Only take the longest valid palindrome
        
        return dp_res[n]

        
        # l = 0
        # ct = 0
        # while l < n-k+1:
        #     flag = False
        #     for r in range(l+k-1,n):
        #         if dp[l][r] == True:
        #             flag = True
        #             ct += 1
        #             print(s[l:r+1])
        #             l = r+1
        #             break
        #     if flag == False:
        #         l += 1
        # return ct
                # Dynamic programming to count max number of non-overlapping palindromes

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