1704. Determine if String Halves Are Alike
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 <= 1000s.lengthis even.sconsists 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;
}
}