花花酱 LeetCode 1726. Tuple with Same Product

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 valids 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) , (10,2,4,5)
(4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)

Example 3:

Input: nums = [2,3,4,6,8,12]
Output: 40

Example 4:

Input: nums = [2,3,5,7]
Output: 0

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 104
• All elements in nums are distinct.

Solution: HashTable

Similar idea to 花花酱 LeetCode 1. Two Sum

Use a hashtable to store all the pair product counts.

Enumerate all possible pairs, increase the answer by the same product counts * 8.

Why time 8? C(4,1) * C(1,1) * C(2,1) * C(1,1)

For pair one AxB, A can be placed at any position in a four tuple, B’s position is then fixed. For another pair CxD, C has two positions to choose from, D is fixed.

Time complexity: O(n^2)
Space complexity: O(n^2)