Description
https://leetcode.com/problems/determine-if-string-halves-are-alike/
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.lengthis even.
- sconsists of uppercase and lowercase letters.
Explanation
check two strings if counts of vowels are the same.
Python Solution
class Solution:
    def halvesAreAlike(self, s: str) -> bool:
        s1 = s[:len(s) // 2]
        s2 = s[len(s) // 2 : ]
        
        count1 = 0
        
        for c in s1:
            if c in ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']:
                count1 += 1
                
        count2 = 0
        for c in s2:
            if c in ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']:
                count2 += 1
                
  
        return count1 == count2- Time Complexity: O(N)
- Space Complexity: O(1)