Solution Explanation: Tuple with Same Product

This problem asks us to find the number of tuples (a, b, c, d) from a given array nums such that a * b = c * d, and a, b, c, and d are distinct.

Approach:

The core idea is to efficiently count pairs with the same product. We iterate through all possible pairs (a, b) from nums and calculate their product. We store these products and their counts in a hash map (or dictionary). Once we have the counts for each unique product, we can easily calculate the number of tuples.

For each product p, if there are n pairs that result in product p, then we can choose any two pairs from these n pairs to form a tuple. The number of ways to choose two pairs from n is given by the combination formula: n! / (2! * (n-2)!) = n * (n-1) / 2. Since each such pair can form 8 distinct tuples (due to the permutations of a, b, c, d), we multiply this combination by 8.

Time Complexity Analysis:

The nested loops iterate through all possible pairs in the array, which takes O(n^2) time, where n is the length of nums. The rest of the operations (hash map lookups and calculations) take constant time per pair, so the overall time complexity is dominated by the nested loops. Therefore, the time complexity is O(n^2).

Space Complexity Analysis:

The hash map stores the products and their counts. In the worst case, every pair will have a unique product, resulting in O(n^2) entries in the hash map. Therefore, the space complexity is O(n^2).

Code Examples:

The provided code examples demonstrate the algorithm in several programming languages. They all follow this same basic structure:

1. Initialization: Create a hash map (dictionary) to store products and their counts.
2. Pair Iteration: Iterate through all possible pairs (a, b) using nested loops.
3. Product Calculation: Compute the product x = a * b.
4. Count Update: Increment the count for the product x in the hash map. If the product is not yet in the map, it's added.
5. Tuple Calculation: After the loops finish, iterate through the hash map's values (counts). For each count v, calculate v * (v - 1) / 2 which represents the number of ways to choose 2 pairs from v pairs with the same product. Then multiply by 8 to account for the permutations of the chosen pairs in a tuple.
6. Return: Sum up the tuple counts for all products and return the total.

The use of bit shifting (ans << 3) in some examples is an optimization for multiplying by 8 (2^3).

Example Usage (Python):

nums = [2, 3, 4, 6]
solution = Solution()
result = solution.tupleSameProduct(nums)  # result will be 8
print(result)
 
nums = [1, 2, 4, 5, 10]
result = solution.tupleSameProduct(nums)  # result will be 16
print(result)
 

This approach efficiently solves the problem by leveraging hash maps to avoid redundant calculations and provide an optimized solution with O(n^2) time and space complexity.