What is the key pattern for solving the 'Check if All Characters Have Equal Number of Occurrences' problem?

The key pattern for this problem is to use a hash table to count character frequencies, followed by checking if all the frequencies are the same.

How do I know if my solution is optimized for large inputs?

Ensure your solution runs in O(n) time complexity and uses space proportional to the number of unique characters in the string.

What are the common pitfalls when solving this problem?

Common pitfalls include failing to check the consistency of frequencies and using inefficient data structures.

Can I solve this problem without using a hash table?

While it's possible to solve this problem with other methods, using a hash table is the most efficient approach for counting and checking frequencies.

What other problems are similar to 'Check if All Characters Have Equal Number of Occurrences'?

Other similar problems involve frequency counting, such as 'Find the Most Frequent Character' or 'Group Anagrams'.

#1941

Easy

auto_awesome哈希·表·结合·string

LeetCode 题解工作台

检查是否所有字符出现次数相同

给你一个字符串 s ，如果 s 是一个好字符串，请你返回 true ，否则请返回 false 。如果 s 中出现过的所有字符的出现次数相同，那么我们称字符串 s 是好字符串。示例 1：输入： s = "abacbc" 输出： true 解释： s 中出现过的字符为 'a'，'b…

哈希表字符串计数

题目描述

给你一个字符串 s ，如果 s 是一个好字符串，请你返回 true ，否则请返回 false 。

如果 s 中出现过的所有字符的出现次数相同，那么我们称字符串 s 是好字符串。

示例 1：

输入：s = "abacbc"
输出：true
解释：s 中出现过的字符为 'a'，'b' 和 'c' 。s 中所有字符均出现 2 次。

示例 2：

输入：s = "aaabb"
输出：false
解释：s 中出现过的字符为 'a' 和 'b' 。
'a' 出现了 3 次，'b' 出现了 2 次，两者出现次数不同。

提示：

1 <= s.length <= 1000
s 只包含小写英文字母。

lightbulb

解题思路

方法一：计数

我们用一个哈希表或者一个长度为 $26$ 的数组 $\textit{cnt}$ 记录字符串 $s$ 中每个字符出现的次数。

接下来遍历 $\textit{cnt}$ 中的每个值，判断所有非零值是否相等即可。

时间复杂度 $O(n)$ ，空间复杂度 $O(|\Sigma|)$ 。其中 $n$ 是字符串 $s$ 的长度；而 $\Sigma$ 是字符集大小，本题中字符集为小写英文字母，因此 $|\Sigma|=26$ 。

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class Solution:
    def areOccurrencesEqual(self, s: str) -> bool:
        return len(set(Counter(s).values())) == 1

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
The candidate should efficiently build and check the frequency table.
question_mark
The candidate should consider edge cases like strings with only one type of character or strings with a large number of unique characters.
question_mark
The candidate should demonstrate an understanding of optimizing with early exits.

warning

常见陷阱

外企场景

error
Forgetting to check the frequency consistency across all characters.
error
Using inefficient data structures that lead to unnecessary computation or memory usage.
error
Not handling strings with varying numbers of unique characters correctly.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Modify the problem to include case-insensitive comparisons.
arrow_right_alt
Allow the string to contain non-English characters.
arrow_right_alt
Optimize the solution to handle very large strings with more efficient space usage.

help

常见问题

外企场景

继续练习

#1897 重新分配字符使所有字符串都相等 #1876 长度为三且各字符不同的子字符串 #2023 连接后等于目标字符串的字符串对