LeetCode 2067: Number of Equal Count Substrings Solution

Master LeetCode problem 2067 (Number of Equal Count Substrings), 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.

2067. Number of Equal Count Substrings

Medium

Hash Table String Sliding Window Counting

Problem Explanation

Explanation:

To solve this problem, we can iterate through all possible substrings and count the occurrence of each character in each substring. For each unique character count, we can keep track of the number of substrings that have the same character count. We can do this by using a hashmap to store the frequencies of character counts.

Algorithm:

Initialize a variable count to store the total number of substrings with equal character counts.
Iterate through all possible substrings using two nested loops.
For each substring, create a hashmap to store the frequencies of character counts.
Calculate the character count by iterating through the substring and updating the frequencies in the hashmap.
Increment the count of substrings with the same character count in the hashmap.
Finally, return the total count of substrings with equal character counts.

Time Complexity: O(n^3) where n is the length of the input string. We have three nested loops to iterate through all possible substrings.

Space Complexity: O(n) for the hashmap to store the frequencies of character counts.

Solution Code

class Solution {
    public int countSubstrings(String s) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            for (int j = i; j < s.length(); j++) {
                Map<Integer, Integer> freq = new HashMap<>();
                for (int k = i; k <= j; k++) {
                    int c = s.charAt(k) - 'a';
                    freq.put(c, freq.getOrDefault(c, 0) + 1);
                }
                int hashCode = freq.hashCode();
                count += freq.getOrDefault(hashCode, 0);
            }
        }
        return count;
    }
}

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