1726. Tuple with Same Product

Problem Description

The problem revolves around finding a specific type of combinations within an array of distinct positive integers. Our goal is to determine how many groups of four elements, (a, b, c, d), exist within the array such that the product of a and b is equal to the product of c and d. There are two constraints in choosing these elements: firstly, all elements a, b, c, and d must be different from each other; secondly, they must be elements from the input array nums. The result we're looking for is the count of these unique tuples within given constraints.

Intuition

When considering the solution to this problem, the key lies in understanding that the equation a * b = c * d suggests pairs (a, b) and (c, d) have the same product, and we need to count the number of such pairs. Given that the pairs (a, b) and (c, d) are interchangeable and the order within a pair doesn't matter (since multiplication is commutative), each product can potentially be involved in several such tuples.

We can approach this problem by using a hash table (or dictionary) to store the counts of products. Here’s a step-by-step breakdown:

1. Iterate over the array using two nested loops to find all possible pairs (a, b) where a != b.
   • For each pair, compute the product and store it in a hash table with the product as the key and the occurrence count as the value.
2. After we have the hash table prepared, we look at the occurrence counts. If a certain product occurs 'n' times, this means there are n pairs of numbers with this product. We can select any 2 pairs out of these n to form the combination mentioned above, which can be calculated using the combination formula C(n, 2), which is n * (n - 1) / 2.
3. For each distinct product with n pairs, we can form combinations that satisfy the equation. Since each combination can produce 8 different tuples (a, b, c, d), (a, b, d, c), (b, a, c, d), etc. (because the pairs can be flipped and their elements can be swapped), we need to multiply the number of possible combinations by 8, which is the same as left-shifting by 3 bits (v * (v - 1) // 2) << 3.
4. Finally, we sum up these results for all different products to get the final answer.

The intuition to use hash tables comes from their ability to track and retrieve the counts of the products efficiently, which is crucial for finding out how many pairs have the same product and subsequently how many tuples can be formed.

Solution Approach

The reference solution provided leverages a defaultdict from Python's collections module, which is a dictionary-like data structure that provides all methods provided by a dictionary but takes a first argument (default_factory) as a default data type for the dictionary. Using int as default_factory, it ensures that if a key is not found in the dictionary, it will be automatically initialized with a default integer value of 0.

Here is a step-by-step explanation of the logic implemented in the solution:

1. A nested for-loop structure is used to generate all unique pairs (a, b) of the input array nums whose indices are (i, j). Python range function excludes the ending index, which avoids using the same element twice and ensures distinctness:

   1for i in range(1, len(nums)):
   2    for j in range(i):
   3        # Calculate the product and update the count.

2. The product x of two distinct numbers nums[i] * nums[j] from the input array is computed:

   1x = nums[i] * nums[j]

3. The product x is used as a key in a cnt defaultdict. The value for this key represents the number of times this product x has been computed and observed so far:

   1cnt[x] += 1

4. After the for-loops have finished executing, each entry in the cnt dictionary will hold a key-value pair, where the key is a distinct product and the value is the count of pairs that resulted in that product.

5. An answer is calculated by iterating through the values of the cnt dictionary. Each value v represents the number of pairs with the same product. As mentioned earlier, C(n, 2) combinations are possible for these pairs, and hence v * (v - 1) // 2 gives us the number of ways to choose 2 pairs:

   1v * (v - 1) // 2

6. Multiplying the number of pair combinations by 8 for each product's tuples, which is achieved by a left shift by 3 (<< 3), provides the count of valid 4-number tuples for that product.

7. Summing all these counts together gives the final result; the total number of tuples (a, b, c, d) that satisfy the condition a * b = c * d:

   1return sum(v * (v - 1) // 2 for v in cnt.values()) << 3

The key algorithmic principle used in this solution is the exploitation of the mathematical properties of combinations and the use of a hash map to efficiently tabulate and retrieve the number of occurrences of each product. This way, the solution maximizes the efficiency of the search process and reduces what could have been a complex problem to a calculation of combinatorial values based on the contents of the dictionary.

Example Walkthrough

Let's consider a small example to illustrate the solution approach.

Suppose we have an input array nums with distinct positive integers [2, 3, 4, 6].

Here's how the solution finds the number of groups of four elements (a, b, c, d) where the product of a * b is equal to the product of c * d, given the constraints that a, b, c, and d are distinct elements from the nums array:

1. We start by iterating over the nums array to find all possible pairs (a, b) using nested loops:

   1cnt = defaultdict(int)
   2for i in range(1, len(nums)):
   3    for j in range(i):
   4        x = nums[i] * nums[j]
   5        cnt[x] += 1

2. As we go forward, we compute the product of each pair and track the count of each product:

   Iteration 1: Pair (3, 2), product 6, count becomes cnt[6] = 1. Iteration 2: Pair (4, 2), product 8, count becomes cnt[8] = 1. Iteration 3: Pair (4, 3), product 12, count becomes cnt[12] = 1. Iteration 4: Pair (6, 2), product 12, and since this product already existed, we increase the count to cnt[12] = 2. Iteration 5: Pair (6, 3), product 18, count becomes cnt[18] = 1. Iteration 6: Pair (6, 4), product 24, count becomes cnt[24] = 1.

3. We have the following cnt products and their occurrence counts:

   1{6: 1, 8: 1, 12: 2, 18: 1, 24: 1}

4. Now, we evaluate the number of tuples for each product. For the product 12, which has 2 occurrences, there can be C(2, 2) = 2 * (2 - 1) / 2 = 1 combination to choose two pairs, which translates into 1 << 3 = 8 different tuples (a, b, c, d).

5. Since other products have only 1 occurrence, they do not contribute to the count of tuples as we need at least 2 pairs - meaning that product 12 is the only one that allows for tuples considering our constraints.

6. The final answer would be the sum of the number of tuples of all products, which in this case is only from the product 12, leading to 8 tuples total.

   1return sum(v * (v - 1) // 2 for v in cnt.values()) << 3
   2# This will return 8 for our example.

In summary, for our example, the number of valid 4-element tuples (a, b, c, d) with the condition a * b = c * d is 8.

Solution Implementation

Time and Space Complexity

The given Python class Solution includes a method tupleSameProduct that calculates the number of tuples (i, j, k, l) such that i < j < k < l and nums[i] * nums[j] = nums[k] * nums[l], with a different ordering of the same numbers not considered as a different tuple.

Time Complexity:

The time complexity of the method is O(n^2) where n is the length of the input nums. This complexity arises as the method uses double nested loops to iterate through pairs of elements in nums. With each loop, the algorithm performs a constant time operation of multiplying the elements and updating a dictionary.

• The outer loop runs (n-1) times (from 1 to n-1).
• The inner loop runs i times for each iteration of the outer loop where i ranges from 0 to n-2. This gives us a total of approximately (n-1)(n/2) pairs, which simplifies to n(n-1)/2 pairs. Considering big O notation, this simplifies further to O(n^2).

Space Complexity:

The space complexity is O(n^2) because in the worst case, the cnt dictionary can grow to store a unique product for every pair of elements in nums. The maximum number of entries in the cnt dictionary is the number of unique pairs of elements which is the same as the total number of pairs (n(n-1)/2), which simplifies to O(n^2) in big O notation when considering the worst-case scenario.

In conclusion:

• The Time Complexity is O(n^2)
• The Space Complexity is O(n^2)

Learn more about how to find time and space complexity quickly using problem constraints.