1726. Tuple with Same Product LeetCode Solution

In this guide, you will get 1726. Tuple with Same Product LeetCode Solution with the best time and space complexity. The solution to Tuple with Same Product problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Tuple with Same Product solution in C++
Tuple with Same Product solution in Java
Tuple with Same Product solution in Python
Additional Resources

1726. Tuple with Same Product LeetCode Solution image

Problem Statement of 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 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.

Complexity Analysis

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

1726. Tuple with Same Product LeetCode Solution in C++

class Solution {
 public:
  int tupleSameProduct(vector<int>& nums) {
    int ans = 0;
    unordered_map<int, int> count;

    for (int i = 0; i < nums.size(); ++i)
      for (int j = 0; j < i; ++j)
        ans += count[nums[i] * nums[j]]++ * 8;

    return ans;
  }
};
/* code provided by PROGIEZ */

1726. Tuple with Same Product LeetCode Solution in Java

class Solution {
  public int tupleSameProduct(int[] nums) {
    int ans = 0;
    Map<Integer, Integer> count = new HashMap<>();

    for (int i = 0; i < nums.length; ++i)
      for (int j = 0; j < i; ++j) {
        final int prod = nums[i] * nums[j];
        ans += count.getOrDefault(prod, 0) * 8;
        count.merge(prod, 1, Integer::sum);
      }

    return ans;
  }
}
// code provided by PROGIEZ

1726. Tuple with Same Product LeetCode Solution in Python

class Solution:
  def tupleSameProduct(self, nums: list[int]) -> int:
    # nums of ways to arrange (a, b) = 2
    # nums of ways to arrange (c, d) = 2
    # nums of ways to arrange (a, b), (c, d) = 2^3 = 8
    ans = 0
    count = collections.Counter()

    for i in range(len(nums)):
      for j in range(i):
        prod = nums[i] * nums[j]
        ans += count[prod] * 8
        count[prod] += 1

    return ans
 # code by PROGIEZ