How can dynamic programming help solve the Maximum Repeating Substring problem?

Dynamic programming helps track the state of consecutive word repetitions in the sequence, ensuring efficient calculation of the maximum k-repeating value.

What is the time complexity of the brute force approach for this problem?

The brute force approach has a time complexity of O(n^2), where n is the length of the sequence, due to checking all possible substrings.

What is the best approach for large inputs in the Maximum Repeating Substring problem?

Dynamic programming or optimized sliding window techniques are better suited for large inputs, offering O(n) time complexity.

Can the solution be optimized further using string matching algorithms?

Yes, advanced string matching algorithms such as KMP or Rabin-Karp could be applied to further optimize the substring search process.

What are the edge cases to consider in this problem?

Edge cases include when the word does not appear in the sequence at all or when the sequence is too short to contain any repetition of the word.

#1668

Easy

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

最大重复子字符串

给你一个字符串 sequence ，如果字符串 word 连续重复 k 次形成的字符串是 sequence 的一个子字符串，那么单词 word 的重复值为 k 。单词 word 的最大重复值是单词 word 在 sequence 中最大的重复值。如果 word 不是 sequence 的子串…

字符串动态规划字符串匹配

题目描述

给你一个字符串 sequence ，如果字符串 word 连续重复 k 次形成的字符串是 sequence 的一个子字符串，那么单词 word 的 重复值为 k 。单词 word 的最大重复值 是单词 word 在 sequence 中最大的重复值。如果 word 不是 sequence 的子串，那么重复值 k 为 0 。

给你一个字符串 sequence 和 word ，请你返回 最大重复值 k 。

示例 1：

输入：sequence = "ababc", word = "ab"
输出：2
解释："abab" 是 "ababc" 的子字符串。

示例 2：

输入：sequence = "ababc", word = "ba"
输出：1
解释："ba" 是 "ababc" 的子字符串，但 "baba" 不是 "ababc" 的子字符串。

示例 3：

输入：sequence = "ababc", word = "ac"
输出：0
解释："ac" 不是 "ababc" 的子字符串。

提示：

1 <= sequence.length <= 100
1 <= word.length <= 100
sequence 和 word 都只包含小写英文字母。

lightbulb

解题思路

方法一：直接枚举

注意到字符串长度不超过 $100$ ，我们直接从大到小枚举 word 的重复次数 $k$ ，判断 word 重复该次数后是否是 sequence 的子串，是则直接返回当前的重复次数 $k$ 。

时间复杂度为 $O(n^2)$ ，其中 $n$ 为 sequence 的长度。

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class Solution:
    def maxRepeating(self, sequence: str, word: str) -> int:
        for k in range(len(sequence) // len(word), -1, -1):
            if word * k in sequence:
                return k

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

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Understanding dynamic programming principles, especially state transitions.
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Ability to recognize the need for optimization when brute force becomes inefficient.
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Knowledge of string matching techniques like sliding windows and substring searches.

warning

常见陷阱

外企场景

error
Overlooking edge cases where the word doesn't appear in the sequence at all.
error
Not optimizing the solution when brute force becomes too slow for larger inputs.
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Misunderstanding how to track repeated substrings efficiently using dynamic programming.

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进阶变体

外企场景

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Consider the case where the sequence contains multiple different words and you need to find the maximum k-repeating value for each.
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Explore optimizing the space complexity of the dynamic programming approach for even larger inputs.
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Extend the problem to allow the word to repeat non-consecutively and calculate the number of total occurrences.

help

常见问题

外企场景

继续练习

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