What is the primary approach to solve Palindrome Partitioning?

Palindrome Partitioning is typically solved by combining dynamic programming to store valid partitions and backtracking to explore all possible ways to split the string into palindromes.

How can you optimize Palindrome Partitioning for larger inputs?

Using memoization for the palindrome check step and dynamic programming for storing intermediate results can significantly optimize the solution, making it more efficient for larger strings.

How does backtracking help in Palindrome Partitioning?

Backtracking helps by generating all possible partitions of the string. It ensures that each partition consists only of palindromic substrings, allowing all valid partitioning options to be explored.

What is the role of dynamic programming in Palindrome Partitioning?

Dynamic programming plays a crucial role by storing intermediate results, specifically whether substrings are palindromes, which makes generating partitions more efficient and avoids redundant computations.

What are some common mistakes when solving Palindrome Partitioning?

Common mistakes include not checking substrings for palindromes before adding them to partitions, failing to use dynamic programming to avoid redundant computations, and not optimizing the algorithm with memoization.

#131

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

分割回文串

给你一个字符串 s ，请你将 s 分割成一些子串，使每个子串都是回文串。返回 s 所有可能的分割方案。示例 1：输入： s = "aab" 输出： [["a","a","b"],["aa","b"]] 示例 2：输入： s = "a" 输出： [["a"]] 提示： 1 s 仅由小写英…

字符串动态规划回溯

题目描述

给你一个字符串 s，请你将 s 分割成一些子串，使每个子串都是 回文串 。返回 s 所有可能的分割方案。

示例 1：

输入：s = "aab"
输出：[["a","a","b"],["aa","b"]]

示例 2：

输入：s = "a"
输出：[["a"]]

提示：

1 <= s.length <= 16
s 仅由小写英文字母组成

lightbulb

解题思路

方法一：预处理 + DFS(回溯)

我们可以使用动态规划，预处理出字符串中的任意子串是否为回文串，即 $f[i][j]$ 表示子串 $s[i..j]$ 是否为回文串。

接下来，我们设计一个函数 $dfs(i)$ ，表示从字符串的第 $i$ 个字符开始，分割成若干回文串，当前分割方案为 $t$ 。

如果 $i=|s|$ ，说明已经分割完成，我们将 $t$ 放入答案数组中，然后返回。

否则，我们可以从 $i$ 开始，从小到大依次枚举结束位置 $j$ ，如果 $s[i..j]$ 是回文串，那么就把 $s[i..j]$ 加入到 $t$ 中，然后继续递归 $dfs(j+1)$ ，回溯的时候要弹出 $s[i..j]$ 。

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

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class Solution:
    def partition(self, s: str) -> List[List[str]]:
        def dfs(i: int):
            if i == n:
                ans.append(t[:])
                return
            for j in range(i, n):
                if f[i][j]:
                    t.append(s[i : j + 1])
                    dfs(j + 1)
                    t.pop()

        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 f[i + 1][j - 1]
        ans = []
        t = []
        dfs(0)
        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
Do you understand how to use dynamic programming with backtracking in palindrome partitioning?
question_mark
Can you efficiently identify palindromes and generate all partitions using state transitions?
question_mark
Will you be able to optimize the solution with memoization to handle larger inputs?

warning

常见陷阱

外企场景

error
Forgetting to check for palindrome substrings before adding them to partitions can result in incorrect outputs.
error
Not using dynamic programming effectively to store intermediate results can lead to redundant computations, slowing down the solution.
error
Failing to optimize the algorithm with memoization can cause performance issues for larger strings, especially with repeated checks.

swap_horiz

进阶变体

外企场景

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Palindrome Partitioning II (min cuts): Find the minimum number of cuts needed for a palindrome partition.
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Partition Strings into Palindromes (non-overlapping): Partition the string with no overlap and each substring being a palindrome.
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Palindrome Partitioning III: Partition the string with the condition of maximizing the number of palindromic partitions.

help

常见问题

外企场景

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