# Leetcode - Determine if String Halves Are Alike Solution

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.

## Solution in Python

```
class Solution:
@staticmethod
def vowelsCount(s):
return sum(1 for i in s if i in {"a","e","i","o","u"})
def halvesAreAlike(self, s: str) -> bool:
s = s.lower()
midpoint = len(s)//2
first_half = s[:midpoint]
second_half = s[midpoint:]
return self.vowelsCount(first_half)==self.vowelsCount(second_half)
```