2183. Count Array Pairs Divisible by K

Problem

Given a 0-indexed integer array nums of length n and an integer k, return **the *number of pairs*** (i, j) *such that:*

• 0 <= i < j <= n - 1 and

• nums[i] * nums[j] is divisible by k.

  Example 1:

Input: nums = [1,2,3,4,5], k = 2
Output: 7
Explanation: 
The 7 pairs of indices whose corresponding products are divisible by 2 are
(0, 1), (0, 3), (1, 2), (1, 3), (1, 4), (2, 3), and (3, 4).
Their products are 2, 4, 6, 8, 10, 12, and 20 respectively.
Other pairs such as (0, 2) and (2, 4) have products 3 and 15 respectively, which are not divisible by 2.    

Example 2:

Input: nums = [1,2,3,4], k = 5
Output: 0
Explanation: There does not exist any pair of indices whose corresponding product is divisible by 5.

  Constraints:

• 1 <= nums.length <= 10^5

• 1 <= nums[i], k <= 10^5

Solution

class Solution {
    public long countPairs(int[] nums, int k) {
        long count = 0L;
        Map<Integer, Long> map = new HashMap<>();
        for (int num : nums) {
            int gd = gcd(num, k);
            int want = k / gd;
            for (Map.Entry<Integer, Long> entry : map.entrySet()) {
                if (entry.getKey() % want == 0) {
                    count += entry.getValue();
                }
            }
            map.put(gd, map.getOrDefault(gd, 0L) + 1L);
        }
        return count;
    }

    private int gcd(int a, int b) {
        if (a > b) {
            return gcd(b, a);
        }
        if (a == 0) {
            return b;
        }
        return gcd(a, b % a);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).