Check if a String Is an Acronym of Words — LeetCode 2828 Python Solution

EasyArrayString
Problem
#2828
Pattern
Hash Map
Reading time
2 min

The problem

Given an array of strings words and a string s, determine if s is an acronym of words. The string s is considered an acronym of words if it can be formed by concatenating the first character of each string in words in order.

Example

Input
words = ["alice","bob","charlie"], s = "abc"
Output
true
Explanation
The first character in the words "alice", "bob", and "charlie" are 'a', 'b', and 'c', respectively. Hence, s = "abc" is the acronym.

Python solution

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

Complexity

MeasureComplexity
TimeO(n)
SpaceO(n) auxiliary

Pattern: Hash Map

Trade memory for time: remember what you have seen so the second pass never happens. LeetCode 2828. Check if a String Is an Acronym of Words is filed here because the reference solution below belongs to the algorithm family this hub collects, even though its LeetCode tags point elsewhere.

The hash map guide has the Python template for the pattern and the 709 LeetCode problems that use it.

Related problems

Frequently asked questions

How hard is LeetCode 2828. Check if a String Is an Acronym of Words?
LeetCode 2828. Check if a String Is an Acronym of Words is rated Easy on LeetCode.
What is the time complexity of LeetCode 2828. Check if a String Is an Acronym of Words?
The Python solution on this page runs in O(n).
What is the space complexity of LeetCode 2828. Check if a String Is an Acronym of Words?
The Python solution on this page uses O(n) auxiliary space.
What topics does LeetCode 2828. Check if a String Is an Acronym of Words cover?
LeetCode 2828. Check if a String Is an Acronym of Words is tagged Array and String on LeetCode.

Stuck on problems like this in a live interview?

Stealth Interview is a desktop app for macOS and Windows. It reads the problem off your screen and returns a working solution with a step-by-step explanation and its time and space complexity — invisible to screen sharing.

Get Stealth Interview