Solution: Determine if String Halves Are Alike

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

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Description:

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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.

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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

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Constraints:

• 2 <= s.length <= 1000
• s.length is even.
• s consists of uppercase and lowercase letters.

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Idea:

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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.

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Javascript Code:

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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
};

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Python Code:

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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

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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;
    }
}

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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;
    }
};

Comments

[bugb]

I think using Map or Object will reduce time to check. 'aeiouAEIOU'.includes always O(n) but using Map it is only O(1)

[Anand-Baraik]

great. thanks for sharing.