<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 Node:
    def __init__(self, len, sfx = None):
        self.len = len
        self.sfx = sfx
        self.next = {}
        
class Solution:
    def maxPalindromes(self, s: str, k: int) -> int:
        curr = empty = Node(0, Node(-1))
        ans_len = ans_idx = 0
        n = len(s)
        dp = [0] * (n + 1)
        for i, c in enumerate(s):
            while i <= curr.len or s[i-1-curr.len] != c:
                curr = curr.sfx
            if c in curr.next:
                curr = curr.next[c]
            else:
                node = curr.next[c] = Node(curr.len+2, empty)
                if node.len > ans_len:
                    ans_len, ans_idx = node.len, i-node.len+1
                if node.len > 1:
                    curr = curr.sfx
                    while i <= curr.len or s[i-1-curr.len] != c:
                        curr = curr.sfx
                    node.sfx = curr.next[c]
                curr = node
            u = curr
            dp[i + 1] = dp[i]
            while u.len >= k:
                dp[i + 1] = max(dp[i + 1], 1 + dp[i - u.len + 1])
                u = u.sfx
        return dp[n]
        
704