1726. Tuple with Same Product

Problem Statement

Medium

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] <= 10⁴
All elements in nums are distinct.

Java

Source file

class Solution {
    public int tupleSameProduct(int[] nums) {
        int n = nums.length, count = 0;
        Map<Integer, Integer> pdtMap = new HashMap<>();
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int pdt = nums[i] * nums[j];
                pdtMap.put(pdt, pdtMap.getOrDefault(pdt, 0) + 1);
            }
        }
        for (Integer freq : pdtMap.values())
            count += freq * (freq - 1) * 4; // nP2 * 2P2 * 2P2
        return count;
    }
}