# LeetCode 1704. Determine if String Halves Are Alike

## 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.length` is even.
• `s` consists 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)