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 <= 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;
}
}