###### tags: `leetcode` `easy` `String` `Counting`
# [1704. Determine if String Halves Are Alike](https://leetcode.com/problems/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`.

## Examples
### 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:

- $2 \leq s.length \leq 1000$
- ``s.length`` is even.
- `s` consists of **uppercase and lowercase** letters.

## Code

int checkVowels(char *c){
    if(*c == 'a')
        return 1;
    if(*c == 'e')
        return 1;
    if(*c == 'i')
        return 1;
    if(*c == 'o')
        return 1;
    if(*c == 'u')
        return 1;
    
    if(*c == 'A')
        return 1;
    if(*c == 'E')
        return 1;
    if(*c == 'I')
        return 1;
    if(*c == 'O')
        return 1;
    if(*c == 'U')
        return 1;
    
    return 0;
}

bool halvesAreAlike(char * s){
    int length = strlen(s), count = 0;
    for(int i = 0 ; i < length/2 ; i++)
        count+=checkVowels(s+i);
    for(int i = length/2 ; i < length ; i++)
        count-=checkVowels(s+i);
    if(!count)
        return true;
    return false;
}

## Complexity

|Space |Time |
|- |- |
|$O(1)$|$O(N)$|

## Result

- Runtime: 0 ms, faster than 100.00%
- Memory Usage: 5.9 MB, less than 31.58%