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)

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