What is the best approach to solving the Count the Number of Special Characters I problem?

The hash table approach is the most efficient, allowing you to track character occurrences and check for special letters in one pass.

How does the hash table approach work in this problem?

By storing the occurrence of each character, you can efficiently check if a character appears both in lowercase and uppercase in a single pass.

What are common pitfalls in solving this problem?

A common mistake is over-complicating the problem with unnecessary iterations or failing to handle edge cases like empty strings.

How does GhostInterview help with this problem?

GhostInterview offers insights into optimal solutions and helps you avoid common pitfalls, improving both accuracy and efficiency.

Can the solution be optimized further for larger inputs?

The hash table approach is already optimal in terms of time complexity for this problem, but you can experiment with different data structures for space efficiency.

#3120

Easy

auto_awesome哈希·表·结合·string

LeetCode 题解工作台

统计特殊字母的数量 I

给你一个字符串 word 。如果 word 中同时存在某个字母的小写形式和大写形式，则称这个字母为特殊字母。返回 word 中特殊字母的数量。示例 1: 输入： word = "aaAbcBC" 输出： 3 解释： word 中的特殊字母是 'a' 、 'b' 和 'c' 。示例 2:…

哈希表字符串

题目描述

给你一个字符串 word。如果 word 中同时存在某个字母的小写形式和大写形式，则称这个字母为 特殊字母 。

返回 word 中 特殊字母 的数量。

示例 1:

输入：word = "aaAbcBC"

输出：3

解释：

word 中的特殊字母是 'a'、'b' 和 'c'。

示例 2:

输入：word = "abc"

输出：0

解释：

word 中不存在大小写形式同时出现的字母。

示例 3:

输入：word = "abBCab"

输出：1

解释：

word 中唯一的特殊字母是 'b'。

提示：

1 <= word.length <= 50
word 仅由小写和大写英文字母组成。

lightbulb

解题思路

方法一：哈希表或数组

我们用一个哈希表或数组 $s$ 来记录字符串 $word$ 中出现的字符。然后遍历 $26$ 个字母，如果小写字母和大写字母都在 $s$ 中出现，则特殊字符的数量加一。

最后返回特殊字符的数量即可。

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class Solution:
    def numberOfSpecialChars(self, word: str) -> int:
        s = set(word)
        return sum(a in s and b in s for a, b in zip(ascii_lowercase, ascii_uppercase))

speed

复杂度分析

指标	值
时间	complexity depends on the approach. The hash table approach is O(n), where n is the length of the string. The brute force method is O(n^2), and the set-based approach also works in O(n), making it efficient for the given constraints. Space complexity is O(1) for the hash table or sets, as the size is limited to the 26 characters in each case.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Focus on solving the problem efficiently within the given constraints.
question_mark
A correct approach with a hash table will demonstrate the ability to optimize with space and time.
question_mark
Look for the candidate's ability to handle edge cases like strings with no uppercase characters.

warning

常见陷阱

外企场景

error
Over-complicating the solution with unnecessary passes through the string.
error
Ignoring edge cases, such as when there are no special characters or when the string is very short.
error
Using inefficient data structures that may lead to excessive time complexity.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Change the input string to include digits or special characters and see if the solution adapts.
arrow_right_alt
Increase the string length to test the performance of the algorithm.
arrow_right_alt
Test the solution with uppercase and lowercase characters mixed in various sequences.

help

常见问题

外企场景

继续练习

#3121 统计特殊字母的数量 II #3137 K 周期字符串需要的最少操作次数 #3138 同位字符串连接的最小长度