What is the solution pattern for 'Check if a String Is an Acronym of Words'?

The problem is an example of using array manipulation and string building, where you concatenate the first letters of the words and compare them to the given string.

How do I optimize this problem for large inputs?

To optimize for large inputs, check the lengths of the words and the acronym string `s` first. If they don't match, return false immediately without concatenating the characters.

What are the common pitfalls in solving this problem?

A common pitfall is failing to check the lengths of `words` and `s` beforehand, leading to unnecessary computations.

What should I do if I encounter an edge case like an empty acronym string?

Handle edge cases by adding checks for empty strings or single-word inputs. Make sure to return false if the lengths do not match.

Can the approach be improved for specific use cases?

Yes, for specific cases such as very large inputs or case-sensitive acronyms, you can optimize the solution by using hash-based or early exit strategies.

#2828

Easy

auto_awesome数组·string

LeetCode 题解工作台

判别首字母缩略词

给你一个字符串数组 words 和一个字符串 s ，请你判断 s 是不是 words 的首字母缩略词。如果可以按顺序串联 words 中每个字符串的第一个字符形成字符串 s ，则认为 s 是 words 的首字母缩略词。例如， "ab" 可以由 ["apple", "banana"] 形成，但…

数组字符串

题目描述

给你一个字符串数组 words 和一个字符串 s ，请你判断 s 是不是 words 的 首字母缩略词 。

如果可以按顺序串联 words 中每个字符串的第一个字符形成字符串 s ，则认为 s 是 words 的首字母缩略词。例如，"ab" 可以由 ["apple", "banana"] 形成，但是无法从 ["bear", "aardvark"] 形成。

如果 s 是 words 的首字母缩略词，返回 true ；否则，返回 false 。

示例 1：

输入：words = ["alice","bob","charlie"], s = "abc"
输出：true
解释：words 中 "alice"、"bob" 和 "charlie" 的第一个字符分别是 'a'、'b' 和 'c'。因此，s = "abc" 是首字母缩略词。

示例 2：

输入：words = ["an","apple"], s = "a"
输出：false
解释：words 中 "an" 和 "apple" 的第一个字符分别是 'a' 和 'a'。
串联这些字符形成的首字母缩略词是 "aa" 。
因此，s = "a" 不是首字母缩略词。

示例 3：

输入：words = ["never","gonna","give","up","on","you"], s = "ngguoy"
输出：true
解释：串联数组 words 中每个字符串的第一个字符，得到字符串 "ngguoy" 。
因此，s = "ngguoy" 是首字母缩略词。

提示：

1 <= words.length <= 100
1 <= words[i].length <= 10
1 <= s.length <= 100
words[i] 和 s 由小写英文字母组成

lightbulb

解题思路

方法一：模拟

我们可以遍历数组 $words$ 中的每个字符串，将其首字母拼接起来，得到一个新的字符串 $t$ ，然后判断 $t$ 是否等于 $s$ 即可。

时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ 。其中 $n$ 是数组 $words$ 的长度。

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class Solution:
    def isAcronym(self, words: List[str], s: str) -> bool:
        return "".join(w[0] for w in words) == s

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
Candidate demonstrates the ability to manipulate arrays and strings efficiently.
question_mark
Candidate should be able to recognize the problem's pattern as an acronym check.
question_mark
Look for an understanding of early exit conditions to avoid unnecessary work.

warning

常见陷阱

外企场景

error
Failing to check the lengths of `s` and `words` upfront, leading to unnecessary string building.
error
Not handling edge cases like when `words` has only one word or when `s` is empty.
error
Not considering the case where the acronym string is longer than `s`, causing unnecessary iterations.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Check if a string is an acronym using both uppercase and lowercase letters.
arrow_right_alt
Implement a version that checks the acronym only for specific words in the list.
arrow_right_alt
Optimize the solution for handling larger datasets efficiently.

help

常见问题

外企场景

继续练习

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