What is the optimal approach for solving Palindrome Partitioning IV?

The optimal approach involves preprocessing palindrome checks in O(n^2) time and then using dynamic programming to solve the problem efficiently in O(n^2) time and space.

How can I preprocess palindromes for efficient substring checking?

You can preprocess palindromes using a 2D dynamic programming table where `dp[i][j]` is `true` if the substring `s[i..j]` is a palindrome. This allows O(1) checks for each substring during state transitions.

What should I consider when optimizing the Palindrome Partitioning IV solution?

Consider optimizing space complexity by reducing the size of the palindrome table and avoiding redundant checks during state transitions.

How can I modify Palindrome Partitioning IV to handle a larger input size?

You can optimize by reducing space complexity and implementing more efficient dynamic programming transitions, while keeping the preprocessing step O(n^2).

What is the pattern behind the Palindrome Partitioning IV problem?

The problem follows the state transition dynamic programming pattern, where subproblems are solved based on earlier solutions and palindromic substrings are preprocessed for efficient checks.

#1745

Hard

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

分割回文串 IV

给你一个字符串 s ，如果可以将它分割成三个非空回文子字符串，那么返回 true ，否则返回 false 。当一个字符串正着读和反着读是一模一样的，就称其为回文字符串。示例 1：输入： s = "abcbdd" 输出： true 解释： "abcbdd" = "a" + "bcb" +…

字符串动态规划

题目描述

给你一个字符串 s ，如果可以将它分割成三个非空回文子字符串，那么返回 true ，否则返回 false 。

当一个字符串正着读和反着读是一模一样的，就称其为 回文字符串 。

示例 1：

输入：s = "abcbdd"
输出：true
解释："abcbdd" = "a" + "bcb" + "dd"，三个子字符串都是回文的。

示例 2：

输入：s = "bcbddxy"
输出：false
解释：s 没办法被分割成 3 个回文子字符串。

提示：

3 <= s.length <= 2000
s 只包含小写英文字母。

lightbulb

解题思路

方法一：动态规划

我们定义 $f[i][j]$ 表示字符串 $s$ 的第 $i$ 个字符到第 $j$ 个字符是否为回文串，初始时 $f[i][j] = \textit{true}$ 。

然后我们可以通过以下的状态转移方程来计算 $f[i][j]$ ：

f[i][j] = \begin{cases} \textit{true}, & \text{if } s[i] = s[j] \text{ and } (i + 1 = j \text{ or } f[i + 1][j - 1]) \\ \textit{false}, & \text{otherwise} \end{cases}

由于 $f[i][j]$ 依赖于 $f[i + 1][j - 1]$ ，因此，我们需要从大到小的顺序枚举 $i$ ，从小到大的顺序枚举 $j$ ，这样才能保证当计算 $f[i][j]$ 时 $f[i + 1][j - 1]$ 已经被计算过。

接下来，我们枚举第一个子串的右端点 $i$ ，第二个子串的右端点 $j$ ，那么第三个子串的左端点可以枚举的范围为 $[j + 1, n - 1]$ ，其中 $n$ 是字符串 $s$ 的长度。如果第一个子串 $s[0..i]$ 、第二个子串 $s[i+1..j]$ 和第三个子串 $s[j+1..n-1]$ 都是回文串，那么我们就找到了一种可行的分割方案，返回 $\textit{true}$ 。

枚举完所有的分割方案后，如果没有找到符合要求的分割方案，那么返回 $\textit{false}$ 。

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

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class Solution:
    def checkPartitioning(self, s: str) -> bool:
        n = len(s)
        f = [[True] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            for j in range(i + 1, n):
                f[i][j] = s[i] == s[j] and (i + 1 == j or f[i + 1][j - 1])
        for i in range(n - 2):
            for j in range(i + 1, n - 1):
                if f[0][i] and f[i + 1][j] and f[j + 1][-1]:
                    return True
        return False

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
Can the candidate efficiently preprocess palindrome checks in O(1) time for each substring?
question_mark
Does the candidate understand the importance of dynamic programming for breaking down the problem into smaller subproblems?
question_mark
Can the candidate discuss trade-offs in time and space complexity for this problem?

warning

常见陷阱

外企场景

error
Forgetting to preprocess palindrome checks, leading to inefficient solutions with redundant checks.
error
Overcomplicating the dynamic programming approach by not leveraging the preprocessed palindrome table.
error
Not correctly handling edge cases, such as strings that cannot be split into three palindromes.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Generalize the problem by allowing a different number of palindromic substrings (e.g., 2 or 4 substrings).
arrow_right_alt
Extend the problem to handle a larger input size or strings with uppercase characters.
arrow_right_alt
Optimize the solution further by reducing the space complexity of the palindrome check table.

help

常见问题

外企场景

继续练习

#1771 由子序列构造的最长回文串的长度 #1668 最大重复子字符串 #1653 使字符串平衡的最少删除次数