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)