<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:
    # Time: O()
    # Space: O(), where n = len(s)
    def maxPalindromes(self, s: str, k: int) -> int:
 
        n = len(s)

        # dp_palindrome[i][j] = 1 if s[i:j] is palindrome
        dp_palindrome = [[0]*(n) for i in range(n)]
        # Single char are palindrome
        for i in range(n):
            dp_palindrome[i][i] = 1
        # Two equal continuous characters are palindrome
        for i in range(n-1):
            dp_palindrome[i][i+1] = (s[i] == s[i+1])
        for length in range(3, n+1):
            for j in range(length-1, n):
                i = j - length + 1
                dp_palindrome[i][j] = dp_palindrome[i+1][j-1] and (s[i] == s[j])

        # dp[i]: maximum number of substrings in s[:i]
        dp = [0]*(n+1)

        for j in range(1, n + 1):
            # Default: no new palindrome ending at j-1
            dp[j] = dp[j - 1]
            
            # Check all possible starting indices i for substrings ending at j-1
            # Ensure that the substring length is at least k
            for i in range(0, j - k + 1):
                if dp_palindrome[i][j - 1]:
                    dp[j] = max(dp[j], dp[i] + 1)

        return dp[n]
3164