How do I approach the Maximum Number of Non-overlapping Palindrome Substrings problem?

Start by identifying all possible palindromic substrings of length at least k, then use dynamic programming and a two-pointer technique to select non-overlapping substrings.

What is the time complexity for this problem?

The time complexity is typically O(n^2) due to palindrome checks and dynamic programming state transitions, though optimizations can be applied.

Can this problem be solved without dynamic programming?

Dynamic programming is highly effective for this problem, but alternative methods may exist, though they may be less efficient.

What should I consider to avoid overlapping palindromic substrings?

Use a two-pointer approach to track the positions of selected substrings and ensure they do not overlap.

How can I optimize the dynamic programming solution for this problem?

Focus on minimizing recomputation by storing previously computed results for palindromic substrings and maintaining state transitions for efficient selection.

#2472

Hard

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

不重叠回文子字符串的最大数目

给你一个字符串 s 和一个正整数 k 。从字符串 s 中选出一组满足下述条件且不重叠的子字符串：每个子字符串的长度至少为 k 。每个子字符串是一个回文串。返回最优方案中能选择的子字符串的最大数目。子字符串是字符串中一个连续的字符序列。示例 1 ：输入： s = "…

双指针字符串动态规划贪心

题目描述

给你一个字符串 s 和一个正整数 k 。

从字符串 s 中选出一组满足下述条件且 不重叠 的子字符串：

每个子字符串的长度至少为 k 。
每个子字符串是一个 回文串 。

返回最优方案中能选择的子字符串的最大数目。

子字符串 是字符串中一个连续的字符序列。

示例 1 ：

输入：s = "abaccdbbd", k = 3
输出：2
解释：可以选择 s = "abaccdbbd" 中斜体加粗的子字符串。"aba" 和 "dbbd" 都是回文，且长度至少为 k = 3 。
可以证明，无法选出两个以上的有效子字符串。

示例 2 ：

输入：s = "adbcda", k = 2
输出：0
解释：字符串中不存在长度至少为 2 的回文子字符串。

提示：

1 <= k <= s.length <= 2000
s 仅由小写英文字母组成

lightbulb

解题思路

方法一：预处理 + 记忆化搜索

预处理字符串 $s$ ，得到 $dp[i][j]$ 表示字符串 $s[i,..j]$ 是否为回文串。

然后定义函数 $dfs(i)$ 表示从字符串 $s[i,..]$ 中选出最多的不重叠回文子串的个数，即：

\begin{aligned} dfs(i) &= \begin{cases} 0, & i \geq n \\ \max\{dfs(i + 1), \max_{j \geq i + k - 1} \{dfs(j + 1) + 1\}\}, & i < n \end{cases} \end{aligned}

时间复杂度 $O(n^2)$ ，空间复杂度 $O(n^2)$ 。其中 $n$ 为字符串 $s$ 的长度。

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class Solution:
    def maxPalindromes(self, s: str, k: int) -> int:
        @cache
        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

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Candidate efficiently applies dynamic programming to solve the problem.
question_mark
Demonstrates understanding of state transitions and two-pointer technique.
question_mark
Can identify and optimize the problem by carefully considering substring overlap.

warning

常见陷阱

外企场景

error
Failing to efficiently check for palindromes, leading to a slow solution.
error
Not ensuring non-overlapping substrings, causing incorrect selections.
error
Overcomplicating the dynamic programming approach without considering time complexity.

swap_horiz

进阶变体

外企场景

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Modify the problem by requiring palindromic substrings to be of an exact length, not just at least k.
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Limit the total number of palindromic substrings to a fixed number.
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Extend the problem to handle uppercase letters or non-ASCII characters.

help

常见问题

外企场景

继续练习

#3302 字典序最小的合法序列 #2486 追加字符以获得子序列 #2522 将字符串分割成值不超过 K 的子字符串