How do you solve Check if a String Is an Acronym of Words on LeetCode?

Check if a String Is an Acronym of Words (LeetCode #2828) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The acronym is formed by the first characters of each word.

What is the brute force solution for Check if a String Is an Acronym of Words?

The brute force approach for Check if a String Is an Acronym of Words is Brute Force, with O(n) time complexity and O(1) space complexity. The brute-force approach involves manually constructing the acronym by concatenating the first characters of each word in the array and then comparing it to the given string. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if a String Is an Acronym of Words?

Check if a String Is an Acronym of Words uses the following concepts: Array, String. The recommended patterns to study are: Array, String.

What are the interview tips for Check if a String Is an Acronym of Words?

Always check for edge cases, like empty strings or mismatched lengths. Explain your thought process clearly while coding. Consider the efficiency of your solution and mention it during the interview.

What are common mistakes when solving Check if a String Is an Acronym of Words?

Not checking if the lengths of words and s are equal before comparison. Confusing string indexing and character access.

#2828

Check if a String Is an Acronym of Words

Easy

ArrayStringArrayString

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution is similar to the brute-force approach but uses a more efficient way to construct the acronym and perform the comparison in a single pass. This reduces unnecessary string concatenation operations.

⚙️

Algorithm

3 steps

1Step 1: Check if the length of words is equal to the length of s. If not, return false.
2Step 2: Loop through each word in the words array and check if the first character matches the corresponding character in s.
3Step 3: If all characters match, return true; otherwise, return false.

solution.py7 lines

1def isAcronym(words, s):
2    if len(words) != len(s):
3        return False
4    for i in range(len(words)):
5        if words[i][0] != s[i]:
6            return False
7    return True

ℹ

Complexity note: The time complexity remains O(n) because we still loop through each word once, but the space complexity is reduced to O(1) as we do not create any additional strings.

1The acronym is formed by the first characters of each word.
2Length mismatch between words and s immediately indicates false.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.