LeetCode 2949: Count Beautiful Substrings II Solution

Master LeetCode problem 2949 (Count Beautiful Substrings II), a hard 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.

2949. Count Beautiful Substrings II

Hard

Hash Table Math String Number Theory Prefix Sum

Problem Explanation

Explanation:

To solve this problem, we can iterate through all possible substrings of the given string s, calculate the number of vowels and consonants in each substring, and then check if the conditions for a beautiful substring are satisfied. We will maintain a count of beautiful substrings that we encounter.

Create a helper function to check if a given substring is beautiful according to the conditions provided.
Iterate through all possible substrings of s.
For each substring, calculate the number of vowels and consonants.
Check if the substring is beautiful using the helper function.
Increment the count of beautiful substrings if the conditions are met.
Return the final count.

Time Complexity:

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

Space Complexity:

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

Solution Code

class Solution {
    public int countBeautifulString(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 substr = s.substring(i, j);
                int vowels = 0, consonants = 0;
                for (char c : substr.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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