tags: `leetcode` `easy` `String` `Counting`

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.

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%

tags: leetcode easy String Counting