1704

1704. Determine if String Halves Are Alike (Easy)

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.

Example 3:

Input: s = "MerryChristmas"
Output: false

Example 4:

Input: s = "AbCdEfGh"
Output: true

 

Constraints:

• 2 <= s.length <= 1000
• s.length is even.
• s consists of uppercase and lowercase letters.

Related Topics:
String

Solution 1.

// OJ: https://leetcode.com/problems/determine-if-string-halves-are-alike/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
    int count(string s) {
        int cnt = 0;
        for (char c : s) {
            c = tolower(c);
            cnt += c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
        }
        return cnt;
    }
public:
    bool halvesAreAlike(string s) {
        return count(s.substr(0, s.size() / 2)) == count(s.substr(s.size() / 2));
    }
};

Or

// OJ: https://leetcode.com/problems/determine-if-string-halves-are-alike/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
    int count(string &s, int begin, int end) {
        int cnt = 0;
        for (int i = begin; i < end; ++i) {
            char c = tolower(s[i]);
            cnt += c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
        }
        return cnt;
    }
public:
    bool halvesAreAlike(string s) {
        return count(s, 0, s.size() / 2) == count(s, s.size() / 2, s.size());
    }
};