Problem Definition & Core Concepts

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. Understanding this problem requires a strong grasp of both combinatorial mathematics and algorithmic efficiency. The naive approach of iterating through all four elements would result in an unacceptable O(N^4) time complexity. By analyzing the properties of multiplication and the fact that all elements are distinct, we can drastically reduce the search space and execution time.

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.

Detailed Mathematical Breakdown

To fully grasp why the optimized approach is so much faster than the naive alternative, it's crucial to understand the mathematical foundations. The problem asks us to find four numbers a, b, c, d such that a * b = c * d. If we think about this in terms of pairs, we are essentially looking for two pairs of numbers, say Pair 1 = (a, b) and Pair 2 = (c, d), that yield the exact same product.

Since the array contains only distinct positive integers, we don't have to worry about a single element being used multiple times in a tuple, except that the definition states a != b != c != d. This is where the distinct elements constraint becomes our biggest advantage. If we pick two elements a and b, and they produce a product P, and we pick another two elements c and d, producing the same product P, because the array elements are completely unique, the set {a, b} and the set {c, d} must be entirely disjoint. They can never share a common number. Why? If they shared a number, say a = c, then a * b = a * d, which implies b = d. But we know elements are distinct, so this scenario is impossible unless we picked the exact same pair twice.

This powerful insight allows us to iterate through all possible pairs (nums[i], nums[j]), calculate their product, and record how many times we've seen this specific product. Once we have the frequencies of each product, calculating the number of valid tuples is a simple matter of combinatorics. If a product appears cnt times, the number of ways to choose 2 pairs out of cnt pairs is cnt * (cnt - 1) / 2. Finally, because the question asks for ordered tuples and we can arrange any valid four numbers in 8 different ways, we multiply the number of combinations by 8.

Optimized C++ Implementation

The solution below leverages an unordered_map to keep track of the frequency of each calculated product from all distinct pairs (nums[i], nums[j]). Since all elements in the input array are distinct, any two pairs that yield the same product must consist of completely different elements. Once the product frequencies are counted, we can iterate through the map to calculate the total valid combinations.

class Solution {
public:
    int tupleSameProduct(vector<int>& nums) {
        uint size = nums.size();
        unordered_map<int, int> mp;

        // Step 1: Calculate frequencies of all pairwise products
        for( int i = 0; i < size; i++ )
            for( int j = i +1; j< size; j++){
                int prod  = nums[i] * nums[j];
                mp[prod]++;
            }
        
        int res = 0;
        // Step 2: Calculate combinations based on identical products
        for( auto& [prod, cnt] : mp ){
            int pairs = (cnt *(cnt-1)) / 2;
            res += 8 * pairs;
        }

        return res;
    }
};

This implementation runs with a time complexity of O(N^2), limited by the double loop generating all initial pairs. The space complexity is also O(N^2), representing the maximum number of unique products we might store in our hash map. By carefully mapping combinations back to permutations (multiplying by 8), we efficiently bypass generating the full combinations sequentially, guaranteeing a highly performant execution under standard constraints.

Frequently Asked Questions (FAQ)