1704. Determine if String Halves Are Alike.py

class Solution(object):
    def halvesAreAlike(self, s):
        """
        :type s: str
        :rtype: bool
        """
#         ctr = 0
#         for i, j in zip(  (range(0, len(s)//2)),  (range(len(s)//2, len(s)))):
#             if s[i] in ('a','e','i','o','u','A', 'E', 'I','O','U'):
#                 ctr-=1
#             if s[j] in ('a','e','i','o','u','A', 'E', 'I','O','U'):
#                 ctr+=1

#         if ctr!=0:
#             return False
#         return True
        l1 = [s[i] for i in range(0, len(s)//2) if s[i] in ('a','e','i','o','u','A', 'E', 'I','O','U')]
        l2 =[s[i] for i in range(len(s)//2, len(s)) if s[i] in ('a','e','i','o','u','A', 'E', 'I','O','U')]
        if len(l1) == len(l2):
            return True
        return False


'''
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.
 

Constraints:

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