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, 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

    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