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1364. Tuple With Same Product

Array Hash Table Counting

Problem - 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.

Solutions

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from collections import defaultdict

class Solution:
    def tupleSameProduct(self, nums: List[int]) -> int:
        count = 0
        hashmap = defaultdict(list)

        for i in range(len(nums)):
            for j in range(i + 1, len(nums)):
                prod = nums[i] * nums[j]
                hashmap[prod].append((nums[i], nums[j]))

        for produ, pair in hashmap.items():
            pairs = len(pair)
            if pairs > 1:
                count += pairs * (pairs - 1) * 4

        return count

Submission Stats:

  • Runtime: 1125 ms (14.13%)
  • Memory: 104.6 MB (14.13%)