# Solving Leetcode Interviews in Seconds with AI: Tuple with Same Product


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1726" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.  

	# Explanation
	Here's a breakdown of the approach and the Python code:

*   **High-Level Approach:**
    *   Calculate the product of each pair of numbers in the input array.
    *   Use a hash map (dictionary in Python) to store the frequency of each product.
    *   Iterate through the hash map, and for each product with frequency `count`, the number of tuples is `count * (count - 1) * 4`.

*   **Complexity:**
    *   Runtime: O(n^2), where n is the length of the input array.  This comes from the nested loops to generate all pairs.
    *   Storage: O(n^2) in the worst case, where n is the length of the input array.  This is because, in the worst case, all products of pairs could be distinct, requiring storage for n*(n-1)/2 distinct entries in the hashmap.

	
	# Code
	```python
	def tupleSameProduct(nums):
    """
    Finds 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.

    Args:
        nums: A list of distinct positive integers.

    Returns:
        The number of valid tuples.
    """
    product_counts = {}
    n = len(nums)
    for i in range(n):
        for j in range(i + 1, n):
            product = nums[i] * nums[j]
            if product in product_counts:
                product_counts[product] += 1
            else:
                product_counts[product] = 1

    count = 0
    for product in product_counts:
        c = product_counts[product]
        count += c * (c - 1) * 8  # Each pair of pairs can be arranged in 8 ways

    return count
	```