LeetCode 2198: Number of Single Divisor Triplets Solution

Master LeetCode problem 2198 (Number of Single Divisor Triplets), 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.

2198. Number of Single Divisor Triplets

Medium

Math

Problem Explanation

Explanation:

The problem asks us to find the number of unique triplets (a, b, c) where a, b, and c are positive integers and a divides b and b divides c.

To solve this problem, we can iterate through all possible triplets and check if the conditions are satisfied. We can do this by generating all divisors for each number and checking if they meet the criteria.

Algorithmic Idea:

Initialize a counter variable count to 0.
Iterate through all possible triplets (a, b, c) where 1 <= a < b < c <= n.
For each triplet, find all divisors of a, b, and c.
Check if a divides b and b divides c. If both conditions are met, increment the count.
Return the final count of valid triplets.

Time Complexity:

The time complexity of this approach is O(n^2 * sqrt(n)) where n is the given integer.

Space Complexity:

The space complexity is O(1) as we are not using any extra space other than a few variables.

Solution Code

class Solution {
    public int countTriplets(int n) {
        int count = 0;
        for (int a = 1; a <= n; a++) {
            for (int b = a + 1; b <= n; b++) {
                for (int c = b + 1; c <= n; c++) {
                    if (b % a == 0 && c % b == 0) {
                        count++;
                    }
                }
            }
        }
        return count;
    }
}

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