Tuple with Same Product - LeetCode solutions

Problem

Given an array nums of distinct positive integers, return **the number of tuples *(a, b, c, d)* such that a * b = c * d where a, b, c, and d are elements of nums, and a != b != c != d.**

Example 1:

Input: nums = [2,3,4,6]
Output: 8
Explanation: There are 8 valid tuples:
(2,6,3,4) , (2,6,4,3) , (6,2,3,4) , (6,2,4,3)
(3,4,2,6) , (4,3,2,6) , (3,4,6,2) , (4,3,6,2)

Example 2:

Input: nums = [1,2,4,5,10]
Output: 16
Explanation: There are 16 valid tuples:
(1,10,2,5) , (1,10,5,2) , (10,1,2,5) , (10,1,5,2)
(2,5,1,10) , (2,5,10,1) , (5,2,1,10) , (5,2,10,1)
(2,10,4,5) , (2,10,5,4) , (10,2,4,5) , (10,2,5,4)
(4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10^4
All elements in nums are distinct.

Solution (Java)

class Solution {
    public int tupleSameProduct(int[] nums) {
        HashMap<Integer, Integer> ab = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                ab.put(nums[i] * nums[j], ab.getOrDefault(nums[i] * nums[j], 0) + 1);
            }
        }
        int count = 0;
        for (Map.Entry<Integer, Integer> entry : ab.entrySet()) {
            int val = entry.getValue();
            count = count + (val * (val - 1)) / 2;
        }
        return count * 8;
    }
}

Explain:

nope.

Complexity:

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