LeetCode 2947: Count Beautiful Substrings I Solution

Master LeetCode problem 2947 (Count Beautiful Substrings I), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2947. Count Beautiful Substrings I

Medium

Hash Table Math String Enumeration Number Theory Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible substrings of the given string s. For each substring, we calculate the number of vowels and consonants. If the count of vowels and consonants are equal and their product is divisible by k, then we increment the count of beautiful substrings.

Algorithm

Initialize a variable count to store the count of beautiful substrings.
Iterate through all possible substrings of the given string s.
For each substring, calculate the count of vowels and consonants.
If the count of vowels and consonants are equal and their product is divisible by k, increment count.
Return the final count as the result.

Time Complexity

The time complexity of this algorithm is O(n^2), where n is the length of the input string s.

Space Complexity

The space complexity of this algorithm is O(1) as we are using a constant amount of extra space.

Solution Code

class Solution {
    public int countBeautifulSubstrings(String s, int k) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            for (int j = i + 1; j <= s.length(); j++) {
                String substring = s.substring(i, j);
                int vowels = 0, consonants = 0;
                for (char c : substring.toCharArray()) {
                    if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
                        vowels++;
                    } else {
                        consonants++;
                    }
                }
                if (vowels == consonants && (vowels * consonants) % k == 0) {
                    count++;
                }
            }
        }
        return count;
    }
}

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