Determine if String Halves Are Alike - Practice Coding Problems

Determine if String Halves Are Alike - Problem

String Counting Easy

Imagine you have a perfectly balanced string of even length, like a word written on a piece of paper that you want to fold in half. Your task is to determine if both halves are "vowel-balanced" - meaning they contain the same number of vowels!

Given a string s of even length, split it into two equal halves: the first half and the second half. Two string halves are considered alike if they contain the same number of vowels.

Vowels include: 'a', 'e', 'i', 'o', 'u' and their uppercase versions 'A', 'E', 'I', 'O', 'U'

Return true if both halves are alike, false otherwise.

Input & Output

example_1.py — Basic Case

$ Input: s = "book"

› Output: true

💡 Note: First half: "bo" has 1 vowel ('o'). Second half: "ok" has 1 vowel ('o'). Since both halves have the same number of vowels (1), return true.

example_2.py — Mixed Case

$ Input: s = "textbook"

› Output: false

💡 Note: First half: "text" has 1 vowel ('e'). Second half: "book" has 2 vowels ('o', 'o'). Since the counts differ (1 ≠ 2), return false.

example_3.py — All Consonants

$ Input: s = "MzXy"

› Output: true

💡 Note: First half: "Mz" has 0 vowels. Second half: "Xy" has 0 vowels. Since both halves have the same number of vowels (0), return true.

Constraints

2 ≤ s.length ≤ 1000
s.length is even
s consists of uppercase and lowercase letters only

Visualization

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Understanding the Visualization

Initialize Scale

Start with a balanced scale (counter = 0)

Add First Half Vowels

Add +1 weight for each vowel in first half

Remove Second Half Vowels

Remove -1 weight for each vowel in second half

Check Balance

Scale is balanced (counter = 0) means equal vowels

Key Takeaway

🎯 Key Insight: Instead of counting vowels separately, we use a balance counter that naturally reaches zero when both halves have equal vowel counts!

Asked in

f Facebook 25 a Amazon 18 G Google 15 ⊞ Microsoft 12

The optimal solution uses a single-pass balance counter approach. We increment a counter for vowels in the first half and decrement for vowels in the second half. If the final balance is zero, the halves have equal vowel counts. Using a HashSet for O(1) vowel lookup makes this solution both time and space efficient with O(n) time and O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Separate Counting)	O(n)	O(1)	Count vowels in each half separately using nested character checking
Single Pass with Balance Counter (Optimal)	O(n)	O(1)	Use a running balance counter: +1 for first half vowels, -1 for second half vowels

Brute Force (Separate Counting) — Algorithm Steps

Split the string into first and second half
For each character in first half, check if it's a vowel by comparing against all vowel characters
Count vowels in first half
Repeat the same process for second half
Compare the two counts and return true if equal

Visualization

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Step-by-Step Walkthrough

Split String

Divide the string into two equal halves

Count First Half

Count vowels in the first half

Count Second Half

Count vowels in the second half

Compare

Compare the two counts

Code -

solution.c — C

bool halvesAreAlike(char* s) {
    char* vowels = "aeiouAEIOU";
    int n = strlen(s);
    int mid = n / 2;
    
    int firstHalfCount = 0;
    int secondHalfCount = 0;
    
    // Count vowels in first half
    for (int i = 0; i < mid; i++) {
        if (strchr(vowels, s[i]) != NULL) {
            firstHalfCount++;
        }
    }
    
    // Count vowels in second half
    for (int i = mid; i < n; i++) {
        if (strchr(vowels, s[i]) != NULL) {
            secondHalfCount++;
        }
    }
    
    return firstHalfCount == secondHalfCount;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We visit each character once, but check against multiple vowels for each character (constant factor)

✓ Linear Growth

Space Complexity

O(1)

Only using a few integer variables to store counts

✓ Linear Space

89.2K Views

Medium Frequency

~8 min Avg. Time

1.8K Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Separate Counting) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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