1364. Tuple With Same Product¶

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1364. Tuple with Same Product

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.

Solution¶

class Solution {
    public int tupleSameProduct(int[] nums) {
        int n = nums.length;
        HashMap<Integer, Integer> map = new HashMap<>();
        int count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                int prod = nums[i] * nums[j];
                count += 8 * map.getOrDefault(prod, 0);
                map.put(prod, map.getOrDefault(prod, 0) + 1);
            }
        }
        return count;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here