---
# System prepended metadata

title: Leetcode 1704. Determine if String Halves Are Alike
tags: [Leetcode]

---

---
###### tags: `Leetcode`
---
# Leetcode 1704. Determine if String Halves Are Alike


[link](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.

#### Constraints:

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

---

We can divide the given string into two parts and then create a helper function to count the vowel characters for each half of the given string.

#### Solution 1
```python=
class Solution:
    def halvesAreAlike(self, s: str) -> bool:
        l = len(s)

        def count(segment):
            count = 0
            vowels = ['a', 'e', 'i', 'o', 'u']
            for i in segment:
                if i.lower() in vowels:
                    count += 1
            
            return count

        return count(s[0:(l//2)]) == count(s[(l//2):l])
```
O(T): O(n)
O(S): O(n)
