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] <= 10^4
• All elements in nums are distinct.

# Optimal Solution

This problem requires us to find the number of tuples (a, b, c, d) such that a * b = c * d, where a, b, c, and d are distinct elements from the input array `nums`. The brute-force approach would involve checking all possible combinations, which is highly inefficient. The optimal solution involves using a hash map to store the frequency of each product of pairs and then calculating the number of valid tuples based on these frequencies.

## Code

```python
from collections import defaultdict

def tuple_same_product(nums):
    product_counts = defaultdict(int)
    count = 0
    n = len(nums)

    for i in range(n):
        for j in range(i + 1, n):
            product = nums[i] * nums[j]
            product_counts[product] += 1

    for product in product_counts:
        freq = product_counts[product]
        count += freq * (freq - 1) * 4

    return count

Example:

nums = [2, 3, 4, 6]
result = tuple_same_product(nums)
print(result)  # Output: 8

nums = [1, 2, 4, 5, 10]
result = tuple_same_product(nums)
print(result)  # Output: 16

Big(O) Run-time Analysis

• The outer loop iterates n times, and the inner loop iterates n - i - 1 times, which is approximately n times in the worst case. Therefore, computing products and updating the hash map takes O(n^2) time.
• Iterating through the hash map and calculating the counts takes O(k) time, where k is the number of unique products. In the worst case, k can be O(n^2), but on average, it will be less.
• Overall, the dominant factor is the nested loop, so the time complexity is O(n^2).

Big(O) Space Usage Analysis

• The space complexity is determined by the product_counts hash map, which stores the frequency of each unique product.
• In the worst case, each pair of numbers in the input array could have a unique product, leading to O(n^2) unique products.
• Therefore, the space complexity is O(n^2).

Edge Cases

1. Empty Input Array:
   • If the input array nums is empty, the function should return 0, as there are no tuples to form.
2. Input Array with One Element:
   • If the input array contains only one element, it's impossible to form a tuple, so the function should return 0.
3. Input Array with Two Elements:
   • If the input array contains two elements, it's impossible to form a tuple, so the function should return 0.
4. Input Array with Three Elements:
   • If the input array contains three elements, it's impossible to form a tuple, so the function should return 0.
5. Large Numbers:
   • If the input array contains very large numbers, the product a * b could exceed the maximum integer value, leading to overflow issues. To handle this, you might consider using larger data types (e.g., long in Java, int64 in Python) or scaling the numbers down.
6. Duplicate Products:
   • The hash map efficiently handles duplicate products by incrementing the count, so no special handling is needed for this case.

Naive Solution

A brute-force solution would involve iterating through all possible combinations of four elements (a, b, c, d) and checking if a * b = c * d. This approach is highly inefficient, especially for large input arrays, due to its high time complexity.

def tuple_same_product_naive(nums):
    count = 0
    n = len(nums)
    for i in range(n):
        for j in range(n):
            if i == j:
                continue
            for k in range(n):
                if k == i or k == j:
                    continue
                for l in range(n):
                    if l == i or l == j or l == k:
                        continue
                    if nums[i] * nums[j] == nums[k] * nums[l]:
                        count += 1
    return count

Time Complexity: O(n^4) Space Complexity: O(1)

This solution is not practical for large input arrays due to its high time complexity.