<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 i in range(n)]
        for i in range(n):
            dp[i][i]=True
        for i in range(n-1,-1,-1):
            for j in range(i+1,n):
                if j==i+1: dp[i][j] = s[i]==s[j]
                else: dp[i][j] = s[i] == s[j] and dp[i+1][j-1]
        @cache 
        def dfs(i):   
            if i==-1:
                return 0
            res = dfs(i-1)
            for j in range(i-k+2):
                if dp[j][i]:
                    res = max(res,dfs(j-1)+1)
            return res
        return dfs(n-1)
3164