<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:
        """
        We start from idx=0 and idx=k-1, and check if we have a palindrome
        or not. If, we find a palindrome, then skip k characters,
        and recurse and try again.
        If not, we just check both possiblities of adding one
        to every end.
        """

        n=len(s)
        
        def checkIfPalindrome(i,j):
            while i<j:
                if s[i]!=s[j]:
                    return False
                i+=1
                j-=1
            return True
        
        @lru_cache(2000)
        def recurse(i, j):
            # if i and j are out of bounds or len substring is
            # greater than k.
            if i >= n or j >= n or (j - i) > k+1:
                return 0
            
            if s[i] == s[j]:
                # if palindrom found, skip k chars, and try again
                # for next characters
                if checkIfPalindrome(i, j):
                    return 1 + recurse(j + 1, j + k)

            # if not, we just check both possiblities of adding one
            # to every end.
            return max(recurse(i, j + 1), recurse(i + 1, j + 1))
        

        return recurse(0, k - 1)
    
                
21200