Determine if String Halves Are Alike

You are given a string s of even length. Split this string into two halves of equal lengths, and let a be the first half and b be the second half.

Two strings are alike if they have the same number of vowels ('a''e''i''o''u''A''E''I''O''U'). Notice that s contains uppercase and lowercase letters.

Return true if a and b are alike. Otherwise, return false.

Example 1:

Input: s = "book"
Output: true
Explanation: a = "bo" and b = "ok". a has 1 vowel and b has 1 vowel. Therefore, they are alike.

Example 2:

Input: s = "textbook"
Output: false
Explanation: a = "text" and b = "book". a has 1 vowel whereas b has 2. Therefore, they are not alike.
Notice that the vowel o is counted twice.

Example 3:

Input: s = "MerryChristmas"
Output: false

Example 4:

Input: s = "AbCdEfGh"
Output: true


  • 2 <= s.length <= 1000
  • s.length is even.
  • s consists of uppercase and lowercase letters.

We can iterate the first half of the s and consider it as a, and similarly, iterate the second half of the s, and consider it as b. And can count the number of vowels in each of the halves.


Step 1: Iterate over the first half of s (i.e., a) and the second half of s (i.e., b). Count the number of vowels respectively.

Step 2: Return if the numbers of vowels equal.

class Solution {
    public boolean halvesAreAlike(String s) {
        int n = s.length();

        String vowels = "aeiouAEIOU";

        int first_half_vowel_count = 0;
        for (int i = 0; i < n / 2; i++) {
            if (vowels.indexOf(s.charAt(i)) != -1) {

        int second_half_vowel_count = 0;
        for (int i = n / 2; i < n; i++) {
            if (vowels.indexOf(s.charAt(i)) != -1) {

        return first_half_vowel_count == second_half_vowel_count;

Complexity Analysis

Let N be the length of s.

  • Time Complexity: O(N), since we need to iterate substring a and b.
  • Space Complexity: O(1), since we do not need extra space. Here we do not take the input s into consideration.

Categories: String

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