1704 - Determine if String Halves Are Alike (Easy)

Problem Link

https://leetcode.com/problems/determine-if-string-halves-are-alike/

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 trueifaandbare 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.

Approach 1: Counting

Well, just count the vowels in each half and compare. Using string_view avoids make a copy of the input.

Clang seems to produce fairly efficient code for isVowel: https://godbolt.org/z/MWra5vb6G -- much shorter than I though it would be and branch-free. @jit posted a nice analysis about the generated code here.

• Time complexity is $O(n)$ and the
• Space complexity is $O(1)$.

C++

Written by @heder

static bool halvesAreAlike(const string_view s) noexcept {
    return vowelCount(s.substr(0, size(s) / 2)) == vowelCount(s.substr(size(s) / 2));
}

static int vowelCount(const string_view s) noexcept {
    return count_if(begin(s), end(s), isVowel);
}

static bool isVowel(char ch) noexcept {
    return
        ch == 'a' || ch == 'e' || ch == 'i'  || ch == 'o' || ch == 'u' ||
        ch == 'A' || ch == 'E' || ch == 'I'  || ch == 'O' || ch == 'U' ;
}

Rust

Written by @wingkwong

impl Solution {
    pub fn halves_are_alike(s: String) -> bool {
        // count the number of vowels in the given string
        let go = |s: &str| s.chars().filter(|&c| "aeiouAEIOU".contains(c)).count();
        // check if the count of the first half is same as that of the second half
        go(&s[..s.len() / 2]) == go(&s[s.len() / 2..])
    }
}