Solution: Determine if String Halves Are Alike

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Leetcode Problem #1704 (Easy): Determine if Strin…


This content originally appeared on DEV Community and was authored by seanpgallivan

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.

Leetcode Problem #1704 (Easy): Determine if String Halves Are Alike

Description:


(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

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.

Examples:

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.

Idea:


(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

This problem is pretty straightforward. The first issue is being able to identify vowels. There are obviously many ways to do this, but this seems like a good place to use some kind of vowel lookup data structure (vowels). Depending on the language, it can be a string, a dictionary, a map, a set, etc.

Then we just need to keep a balance counter (ans) and iterate through both halves of the input string (S) and increment ans whenever the first half has a vowel and decrement ans whenever the second half has a vowel.

Once we're done, we can just return ans == 0 to determine if the two string halves are balanced.

Javascript Code:


(Jump to: Problem Description || Solution Idea)

const vowels = "aeiouAEIOU"

var halvesAreAlike = function(S) {
    let mid = S.length / 2, ans = 0
    for (let i = 0, j = mid; i < mid; i++, j++)
        ans += vowels.includes(S.charAt(i)) - vowels.includes(S.charAt(j))
    return ans === 0
};

Python Code:


(Jump to: Problem Description || Solution Idea)

vowels = "aeiouAEIOU"

class Solution:
    def halvesAreAlike(self, S: str) -> bool:
        mid, ans = len(S) // 2, 0
        for i in range(mid):
            if S[i] in vowels: ans += 1
            if S[mid+i] in vowels: ans -=1
        return ans == 0

Java Code:


(Jump to: Problem Description || Solution Idea)

class Solution {
    String vowels = "aeiouAEIOU";

    public boolean halvesAreAlike(String S) {
        int mid = S.length() / 2, ans = 0;
        for (int i = 0, j = mid; i < mid; i++, j++) {
            if (vowels.indexOf(S.charAt(i)) >= 0) ans++;
            if (vowels.indexOf(S.charAt(j)) >= 0) ans--;
        }
        return ans == 0;
    }
}

C++ Code:


(Jump to: Problem Description || Solution Idea)

string vowels = "aeiouAEIOU";

class Solution {
public:
    bool halvesAreAlike(string S) {
        int mid = S.size() / 2, ans = 0;
        for (int i = 0, j = mid; i < mid; i++, j++) {
            if (~vowels.find(S[i])) ans++;
            if (~vowels.find(S[j])) ans--;
        }
        return ans == 0;
    }
};


This content originally appeared on DEV Community and was authored by seanpgallivan


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