Determine if String Halves Are Alike Leetcode Solution
In this tutorial, we will solve a leetcode problem, determining if string halves are alike.
Task:
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.
Solution:
class Solution:
def halvesAreAlike(self, s: str) -> bool:
lst = ['a','e','i','o','u','A','E','I','O','U']
n = len(s)
m, n = s[:n//2], s[n//2:]
count, count1 = 0, 0
for i in range(len(m)):
if m[i] in lst:
count += 1
if n[i] in lst:
count1 += 1
return count == count1
Steps:
step1: In this problem first, we are going to initiate a list containing all the vowels both uppercase and lowercase.
step2: Now, we declared two strings m and n, which are equal halves of string s.
step3: Now, we declare two more variables count and count1 which then count the number of vowels in each half of the string s.
step4: Then, we loop into these two strings and checks if any of the characters in m & n is in lst, if it is then we increment count & count1 and then we return count == count1 as bool.
