How do you solve Determine if String Halves Are Alike on LeetCode?

Determine if String Halves Are Alike (LeetCode #1704) 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: Both halves must have the same number of vowels.

What is the brute force solution for Determine if String Halves Are Alike?

The brute force approach for Determine if String Halves Are Alike is Brute Force, with O(n) time complexity and O(1) space complexity. The brute-force approach involves counting the vowels in each half of the string separately. This is straightforward but inefficient, especially for longer strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Determine if String Halves Are Alike?

Determine if String Halves Are Alike uses the following concepts: String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Determine if String Halves Are Alike?

Always clarify the constraints and edge cases. Think about how to optimize your solution during the interview. Practice explaining your thought process clearly.

What are common mistakes when solving Determine if String Halves Are Alike?

Not considering both uppercase and lowercase vowels. Failing to handle the string length properly.

#1704

Determine if String Halves Are Alike

Easy

StringCountingHash MapArray

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 uses a single pass to count the vowels in both halves simultaneously, which reduces the number of iterations needed.

⚙️

Algorithm

5 steps

1Step 1: Initialize a vowel count variable.
2Step 2: Loop through the first half and the second half of the string simultaneously.
3Step 3: For each character, check if it's a vowel and update the count accordingly.
4Step 4: After the loop, check if the counts are equal.
5Step 5: Return true if they are equal, otherwise return false.

solution.py12 lines

1# Full working Python code
2
3def halvesAreAlike(s):
4    n = len(s)
5    count = 0
6    vowels = 'aeiouAEIOU'
7    for i in range(n // 2):
8        if s[i] in vowels:
9            count += 1
10        if s[n - 1 - i] in vowels:
11            count -= 1
12    return count == 0

ℹ

Complexity note: The time complexity remains O(n) as we still iterate through half of the string, but we only use a single variable for counting, making the space complexity O(1).

1Both halves must have the same number of vowels.
2Using a single pass can significantly optimize the solution.

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