LeetCode #1704 — EASY

Determine if String Halves Are Alike

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

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.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"book"

Example 2

"textbook"

Step 02

Core Insight

What unlocks the optimal approach

Create a function that checks if a character is a vowel, either uppercase or lowercase.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1704: Determine if String Halves Are Alike
class Solution {
    public boolean halvesAreAlike(String s) {
        Set<Character> vowels = Set.of('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U');
        int n = s.length() >> 1;
        int cnt = 0;
        for (int i = 0; i < n; ++i) {
            cnt += vowels.contains(s.charAt(i)) ? 1 : 0;
            cnt -= vowels.contains(s.charAt(i + n)) ? 1 : 0;
        }
        return cnt == 0;
    }
}

// Accepted solution for LeetCode #1704: Determine if String Halves Are Alike
func halvesAreAlike(s string) bool {
	vowels := map[byte]bool{}
	for _, c := range "aeiouAEIOU" {
		vowels[byte(c)] = true
	}
	cnt, n := 0, len(s)>>1
	for i := 0; i < n; i++ {
		if vowels[s[i]] {
			cnt++
		}
		if vowels[s[i+n]] {
			cnt--
		}
	}
	return cnt == 0
}

# Accepted solution for LeetCode #1704: Determine if String Halves Are Alike
class Solution:
    def halvesAreAlike(self, s: str) -> bool:
        cnt, n = 0, len(s) >> 1
        vowels = set('aeiouAEIOU')
        for i in range(n):
            cnt += s[i] in vowels
            cnt -= s[i + n] in vowels
        return cnt == 0

// Accepted solution for LeetCode #1704: Determine if String Halves Are Alike
impl Solution {
    pub fn halves_are_alike(s: String) -> bool {
        let n = s.len() / 2;
        let vowels: std::collections::HashSet<char> = "aeiouAEIOU".chars().collect();
        let mut cnt = 0;

        for i in 0..n {
            if vowels.contains(&s.chars().nth(i).unwrap()) {
                cnt += 1;
            }
            if vowels.contains(&s.chars().nth(i + n).unwrap()) {
                cnt -= 1;
            }
        }

        cnt == 0
    }
}

// Accepted solution for LeetCode #1704: Determine if String Halves Are Alike
function halvesAreAlike(s: string): boolean {
    const vowels = new Set('aeiouAEIOU'.split(''));
    let cnt = 0;
    const n = s.length >> 1;
    for (let i = 0; i < n; ++i) {
        cnt += vowels.has(s[i]) ? 1 : 0;
        cnt -= vowels.has(s[n + i]) ? 1 : 0;
    }
    return cnt === 0;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.