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Tuple with Same Product

Difficulty: Medium

Problem Description

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.

Key Insights

  • The problem involves finding pairs of products from distinct elements in the array.
  • Since all elements are distinct, we can use a hash map to count the frequency of each product formed by pairs.
  • Each product can correspond to multiple pairs, and we need to calculate the combinations of these pairs to form valid tuples.

Space and Time Complexity

Time Complexity: O(n^2) - We need to check all pairs of elements to calculate products. Space Complexity: O(n^2) - The hash map can potentially store products of all pairs.

Solution

To solve this problem, we can use a hash map to store the count of each unique product formed by pairs of elements in the array. For each product, we can calculate how many pairs produce the same product and then use combinatorial math to determine the number of valid tuples that can be formed. The approach involves:

  1. Iterating through all pairs of elements in the array and calculating their products.
  2. Storing the counts of these products in a hash map.
  3. For each product that has more than one pair, calculating the number of valid tuples using the formula for combinations.

Code Solutions

from collections import defaultdict

def tupleSameProduct(nums):
    product_count = defaultdict(int)
    n = len(nums)
    
    # Calculate all products of pairs
    for i in range(n):
        for j in range(i + 1, n):
            product = nums[i] * nums[j]
            product_count[product] += 1
            
    count = 0
    
    # Calculate the number of valid tuples
    for pairs in product_count.values():
        if pairs > 1:
            count += pairs * (pairs - 1) * 2  # Each pair can form two tuples (a,b) and (c,d)
    
    return count
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