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Easy
1207
Weekly Contest 221 Q1
String
Counting

1704. Determine if String Halves Are Alike

中文文档

Description

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.

Solutions

Solution 1: Counting

Traverse the string. If the number of vowels in the first half of the string is equal to the number of vowels in the second half, return true. Otherwise, return false.

The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(C)$, where $C$ is the number of vowel characters.

Python3

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

Java

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;
    }
}

C++

class Solution {
public:
    bool halvesAreAlike(string s) {
        unordered_set<char> vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
        int cnt = 0, n = s.size() / 2;
        for (int i = 0; i < n; ++i) {
            cnt += vowels.count(s[i]);
            cnt -= vowels.count(s[i + n]);
        }
        return cnt == 0;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}

JavaScript

/**
 * @param {string} s
 * @return {boolean}
 */
var halvesAreAlike = function (s) {
    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]);
        cnt -= vowels.has(s[n + i]);
    }
    return cnt === 0;
};

PHP

class Solution {
    /**
     * @param String $s
     * @return Boolean
     */
    function halvesAreAlike($s) {
        $n = strlen($s) / 2;
        $vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'];
        $cnt = 0;

        for ($i = 0; $i < $n; $i++) {
            if (in_array($s[$i], $vowels)) {
                $cnt++;
            }
            if (in_array($s[$i + $n], $vowels)) {
                $cnt--;
            }
        }

        return $cnt == 0;
    }
}