Leetcode #2472: Maximum Number of Non-overlapping Palindrome Substrings
In this guide, we solve Leetcode #2472 Maximum Number of Non-overlapping Palindrome Substrings in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a string s and a positive integer k. Select a set of non-overlapping substrings from the string s that satisfy the following conditions: The length of each substring is at least k.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Greedy, Two Pointers, String, Dynamic Programming
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: s = "abaccdbbd", k = 3
Output: 2
Explanation: We can select the substrings underlined in s = "abaccdbbd". Both "aba" and "dbbd" 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.
Python Solution
class Solution:
def maxPalindromes(self, s: str, k: int) -> int:
def dfs(i):
if i >= n:
return 0
ans = dfs(i + 1)
for j in range(i + k - 1, n):
if dp[i][j]:
ans = max(ans, 1 + dfs(j + 1))
return ans
n = len(s)
dp = [[True] * n for _ in range(n)]
for i in range(n - 1, -1, -1):
for j in range(i + 1, n):
dp[i][j] = s[i] == s[j] and dp[i + 1][j - 1]
ans = dfs(0)
dfs.cache_clear()
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.