Problem URL: Tuple with same product

We will first count the product of 2 elements to a hashmap, where the key will be product. Then we iterate over each product, if it is more than 1, then we calculate the number of tuple by (n*(n-1)/2)*8.

Each product value represents 2 tuples, which means there are 2 possibilities in permutation. Meanwhile, each tuple can has 2 possibilities of permutation. The total is 222 = 8 possibilities. Also, we need to select 2 tuples in n candidates, which is n(n-1)/2. Putting everthing together, it is (n(n-1)/2)*8.

class Solution:
    def tupleSameProduct(self, nums: List[int]) -> int:
        lookup = collections.defaultdict(int)
        for i in range(len(nums)-1):
            for j in range(i+1, len(nums)):
                lookup[nums[i]*nums[j]] += 1

        res = 0
        for key, val in lookup.items():
            if val == 1:
                continue
            res += ((val*(val-1))//2)*8

        return res

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