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Formatted question description: https://leetcode.ca/all/1704.html

1704. Determine if String Halves Are Alike (Easy)

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.

Related Topics:
String

Solution 1.

// OJ: https://leetcode.com/problems/determine-if-string-halves-are-alike/
// Time: O(N)
// Space: O(N)
class Solution {
    int count(string s) {
        int cnt = 0;
        for (char c : s) {
            c = tolower(c);
            cnt += c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
        }
        return cnt;
    }
public:
    bool halvesAreAlike(string s) {
        return count(s.substr(0, s.size() / 2)) == count(s.substr(s.size() / 2));
    }
};

Or

// OJ: https://leetcode.com/problems/determine-if-string-halves-are-alike/
// Time: O(N)
// Space: O(1)
class Solution {
    int count(string &s, int begin, int end) {
        int cnt = 0;
        for (int i = begin; i < end; ++i) {
            char c = tolower(s[i]);
            cnt += c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
        }
        return cnt;
    }
public:
    bool halvesAreAlike(string s) {
        return count(s, 0, s.size() / 2) == count(s, s.size() / 2, s.size());
    }
};

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public boolean halvesAreAlike(String s) {
          s = s.toLowerCase();
          int length = s.length();
          int halfLength = length / 2;
          int vowels1 = 0, vowels2 = 0;
          for (int i = 0; i < halfLength; i++) {
              char c1 = s.charAt(i), c2 = s.charAt(i + halfLength);
              if (isVowel(c1))
                  vowels1++;
              if (isVowel(c2))
                  vowels2++;
          }
          return vowels1 == vowels2;
      }
  
      public boolean isVowel(char c) {
          return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
      }
  }
  
  ############
  
  class Solution {
      private static final Set<Character> VOWELS
          = Set.of('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U');
  
      public boolean halvesAreAlike(String s) {
          int cnt = 0, n = s.length() >> 1;
          for (int i = 0; i < n; ++i) {
              cnt += VOWELS.contains(s.charAt(i)) ? 1 : 0;
              cnt -= VOWELS.contains(s.charAt(i + n)) ? 1 : 0;
          }
          return cnt == 0;
      }
  }
  

• // OJ: https://leetcode.com/problems/determine-if-string-halves-are-alike/
  // Time: O(N)
  // Space: O(N)
  class Solution {
      int count(string s) {
          int cnt = 0;
          for (char c : s) {
              c = tolower(c);
              cnt += c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
          }
          return cnt;
      }
  public:
      bool halvesAreAlike(string s) {
          return count(s.substr(0, s.size() / 2)) == count(s.substr(s.size() / 2));
      }
  };
  

• class Solution:
      def halvesAreAlike(self, s: str) -> bool:
          cnt, n = 0, len(s) >> 1
          vowels = set('aeiouAEIOU')
          for i in range(n):
              cnt += s[i] in vowels
              cnt -= s[i + n] in vowels
          return cnt == 0
  
  ############
  
  # 1704. Determine if String Halves Are Alike
  # https://leetcode.com/problems/determine-if-string-halves-are-alike/
  
  class Solution:
      def halvesAreAlike(self, s: str) -> bool:
          half = len(s) // 2
  
          def count(s):
              return sum(c in "aeiouAEIOU" for c in s)
          
          return count(s[:half]) == count(s[half:])
          
          
  

• func halvesAreAlike(s string) bool {
  	vowels := map[byte]bool{}
  	for _, c := range "aeiouAEIOU" {
  		vowels[byte(c)] = true
  	}
  	cnt, n := 0, len(s)>>1
  	for i := 0; i < n; i++ {
  		if vowels[s[i]] {
  			cnt++
  		}
  		if vowels[s[i+n]] {
  			cnt--
  		}
  	}
  	return cnt == 0
  }
  

• function halvesAreAlike(s: string): boolean {
      const set = new Set(['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']);
      const n = s.length >> 1;
      let count = 0;
      for (let i = 0; i < n; i++) {
          set.has(s[i]) && count++;
          set.has(s[n + i]) && count--;
      }
      return count === 0;
  }
  
  

• use std::collections::HashSet;
  impl Solution {
      pub fn halves_are_alike(s: String) -> bool {
          let set: HashSet<&u8> = [b'a', b'e', b'i', b'o', b'u', b'A', b'E', b'I', b'O', b'U']
              .into_iter()
              .collect();
          let s = s.as_bytes();
          let n = s.len() >> 1;
          let mut count = 0;
          for i in 0..n {
              if set.contains(&s[i]) {
                  count += 1;
              }
              if set.contains(&s[n + i]) {
                  count -= 1;
              }
          }
          count == 0
      }
  }
  
  

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