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 <= 1000s.lengthis even.sconsists of uppercase and lowercase letters.
Related Topics:
String
// 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());
}
};