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Tuple with Same Product

Medium

Asked by:

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Topics:

Arrays

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.

Try coding it in LeetCode

Tuple with Same Product

Medium

Asked by:

35 views

Topics:

Arrays

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected range of values within the input array? Can the array contain negative numbers, zeros, or floating-point numbers?
What is the expected size of the input array? Is there a maximum size I should consider?
If no tuples with the same product exist, what should I return? Should I return 0, null, or an empty list/array?
Are duplicate numbers allowed in the input array? If so, how should I handle them in forming the tuples?
Does the order of the numbers within a tuple matter? For example, is (a, b, c, d) considered the same as (b, a, c, d)?

Brute Force Solution

Approach

The brute force approach to this problem involves checking every possible combination of numbers to see if they form tuples with the same product. It's like trying every possible lottery ticket to see if it wins. We don't optimize, we just try everything.

Here's how the algorithm would work step-by-step:

Start by picking two numbers from the list.
Multiply those two numbers together to find their product.
Now, go through every other pair of numbers in the list.
For each of these other pairs, multiply them together as well.
If the product of this new pair is the same as the product of the first pair, then you have found a match!
Count this match.
Repeat this process for every possible initial pair of numbers from the list.
Once you've checked all the pairs, you will have counted all the possible matching tuples.

Code Implementation

def tuple_with_same_product_brute_force(numbers):
    count_of_matching_tuples = 0

    # Iterate through each possible first pair
    for first_index in range(len(numbers)):
        for second_index in range(first_index + 1, len(numbers)):

            first_product = numbers[first_index] * numbers[second_index]

            # Iterate through each possible second pair
            for third_index in range(len(numbers)):
                for fourth_index in range(third_index + 1, len(numbers)):

                    # Ensure that the indices are all distinct
                    if (first_index != third_index and
                            first_index != fourth_index and
                            second_index != third_index and
                            second_index != fourth_index):

                        second_product = numbers[third_index] * numbers[fourth_index]

                        # Check if the products are equal
                        if first_product == second_product:
                            #We found a matching tuple
                            count_of_matching_tuples += 1

    # Each tuple (a, b, c, d) can also be (c, d, a, b)
    # so we must multiply by 2
    return count_of_matching_tuples * 2

Big(O) Analysis

Time Complexity

O(n^4) – The algorithm iterates through all possible pairs of numbers in the input array of size n. This involves selecting two numbers, which can be done in approximately n^2 ways (nested loops). For each of these initial pairs, the algorithm then iterates through all *other* possible pairs, which again takes approximately n^2 time. Therefore, the total time complexity is approximately n^2 * n^2, which simplifies to O(n^4).

Space Complexity

O(1) – The brute force approach described only uses a few constant space variables for storing products and counters. No additional data structures like arrays or hash maps are created that scale with the input size, N, which represents the number of elements in the input list. Therefore, the auxiliary space used remains constant irrespective of the input size. The space complexity is thus O(1).

Optimal Solution

Approach

The trick here is to realize we don't need to check every single combination of numbers. Instead, we can focus on the products that result from multiplying pairs of numbers, and count how many times each product appears. These counts will help us quickly find valid tuples.

Here's how the algorithm would work step-by-step:

For every possible pair of numbers in the list, calculate their product.
Keep track of how many times each unique product shows up. Think of it like creating a product tally sheet.
For each product that shows up more than once, figure out how many different ways you can pick two pairs that result in that same product. The number of ways to pick two pairs is based on the number of times the product shows up.
Multiply the number of ways to pick two pairs by 8, because for any set of four numbers that form the two pairs, there are always 8 different valid tuple arrangements.
Add up all these results to get the total number of tuples.

Code Implementation

def tuple_same_product(nums):
    product_counts = {}
    number_of_tuples = 0

    # Count occurrences of each product.
    for i in range(len(nums)):
        for j in range(i + 1, len(nums)):
            product = nums[i] * nums[j]
            product_counts[product] = product_counts.get(product, 0) + 1

    # Iterate through the counts to find pairs of pairs.
    for product in product_counts:
        product_count = product_counts[product]

        # We only care if a product shows up more than once
        if product_count > 1:

            # Number of ways to pick 2 pairs with same product.
            number_of_pairs = product_count * (product_count - 1) // 2
            
            # Each set of 4 numbers has 8 possible tuples.
            number_of_tuples += number_of_pairs * 8

    return number_of_tuples

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible pairs of numbers in the input array of size n to calculate their product. This involves a nested loop where, for each element, the algorithm considers all subsequent elements to form pairs. Calculating the product for each pair takes constant time, but the number of pairs is proportional to n * (n-1) / 2. Therefore, the dominant operation is calculating the products of all possible pairs which leads to approximately n²/2 operations, which simplifies to O(n²).

Space Complexity

O(N^2) – The dominant space usage comes from the product tally sheet, which effectively stores the counts of products formed by pairs of numbers. Since we are calculating the product of every possible pair of numbers from the input list of size N, the number of possible pairs is N * (N-1) / 2 which is approximately N^2/2. Therefore, the tally sheet could potentially store up to O(N^2) unique products and their counts. This leads to an auxiliary space requirement of O(N^2).

Edge Cases

Case	How to Handle
Empty input array	Return 0 immediately as there are no tuples to form.
Input array with fewer than 4 elements	Return 0 as at least 4 elements are required to form two pairs.
Input array with very large numbers leading to potential integer overflow during multiplication	Use a data type that can accommodate larger numbers or modulo the multiplication result to prevent overflow.
Input array containing zero	Any product involving zero will always be zero; handle zero values separately to avoid invalid combinations.
Input array with all elements being 1	Calculate the number of possible pairs directly using combinatorics.
Large input array with many pairs having the same product	Ensure the HashMap storing products has sufficient capacity and the collision handling is efficient.
Input array containing duplicate numbers that form valid tuples	The solution should correctly count valid tuple combinations even with duplicate numbers, without reusing the same index.
The array contains negative numbers	The product map should consider both positive and negative products.

Empty input array

How to Handle:

Return 0 immediately as there are no tuples to form.

Input array with fewer than 4 elements

How to Handle:

Return 0 as at least 4 elements are required to form two pairs.

Input array with very large numbers leading to potential integer overflow during multiplication

How to Handle:

Use a data type that can accommodate larger numbers or modulo the multiplication result to prevent overflow.

Input array containing zero

How to Handle:

Any product involving zero will always be zero; handle zero values separately to avoid invalid combinations.

Input array with all elements being 1

How to Handle:

Calculate the number of possible pairs directly using combinatorics.

Large input array with many pairs having the same product

How to Handle:

Ensure the HashMap storing products has sufficient capacity and the collision handling is efficient.

Input array containing duplicate numbers that form valid tuples

How to Handle:

The solution should correctly count valid tuple combinations even with duplicate numbers, without reusing the same index.

The array contains negative numbers

How to Handle:

The product map should consider both positive and negative products.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected range of values within the input array? Can the array contain negative numbers, zeros, or floating-point numbers?
What is the expected size of the input array? Is there a maximum size I should consider?
If no tuples with the same product exist, what should I return? Should I return 0, null, or an empty list/array?
Are duplicate numbers allowed in the input array? If so, how should I handle them in forming the tuples?
Does the order of the numbers within a tuple matter? For example, is (a, b, c, d) considered the same as (b, a, c, d)?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by picking two numbers from the list.
Multiply those two numbers together to find their product.
Now, go through every other pair of numbers in the list.
For each of these other pairs, multiply them together as well.
If the product of this new pair is the same as the product of the first pair, then you have found a match!
Count this match.
Repeat this process for every possible initial pair of numbers from the list.
Once you've checked all the pairs, you will have counted all the possible matching tuples.

Code Implementation

def tuple_with_same_product_brute_force(numbers):
    count_of_matching_tuples = 0

    # Iterate through each possible first pair
    for first_index in range(len(numbers)):
        for second_index in range(first_index + 1, len(numbers)):

            first_product = numbers[first_index] * numbers[second_index]

            # Iterate through each possible second pair
            for third_index in range(len(numbers)):
                for fourth_index in range(third_index + 1, len(numbers)):

                    # Ensure that the indices are all distinct
                    if (first_index != third_index and
                            first_index != fourth_index and
                            second_index != third_index and
                            second_index != fourth_index):

                        second_product = numbers[third_index] * numbers[fourth_index]

                        # Check if the products are equal
                        if first_product == second_product:
                            #We found a matching tuple
                            count_of_matching_tuples += 1

    # Each tuple (a, b, c, d) can also be (c, d, a, b)
    # so we must multiply by 2
    return count_of_matching_tuples * 2

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

For every possible pair of numbers in the list, calculate their product.
Keep track of how many times each unique product shows up. Think of it like creating a product tally sheet.
For each product that shows up more than once, figure out how many different ways you can pick two pairs that result in that same product. The number of ways to pick two pairs is based on the number of times the product shows up.
Multiply the number of ways to pick two pairs by 8, because for any set of four numbers that form the two pairs, there are always 8 different valid tuple arrangements.
Add up all these results to get the total number of tuples.

Code Implementation

def tuple_same_product(nums):
    product_counts = {}
    number_of_tuples = 0

    # Count occurrences of each product.
    for i in range(len(nums)):
        for j in range(i + 1, len(nums)):
            product = nums[i] * nums[j]
            product_counts[product] = product_counts.get(product, 0) + 1

    # Iterate through the counts to find pairs of pairs.
    for product in product_counts:
        product_count = product_counts[product]

        # We only care if a product shows up more than once
        if product_count > 1:

            # Number of ways to pick 2 pairs with same product.
            number_of_pairs = product_count * (product_count - 1) // 2
            
            # Each set of 4 numbers has 8 possible tuples.
            number_of_tuples += number_of_pairs * 8

    return number_of_tuples

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 immediately as there are no tuples to form.
Input array with fewer than 4 elements	Return 0 as at least 4 elements are required to form two pairs.
Input array with very large numbers leading to potential integer overflow during multiplication	Use a data type that can accommodate larger numbers or modulo the multiplication result to prevent overflow.
Input array containing zero	Any product involving zero will always be zero; handle zero values separately to avoid invalid combinations.
Input array with all elements being 1	Calculate the number of possible pairs directly using combinatorics.
Large input array with many pairs having the same product	Ensure the HashMap storing products has sufficient capacity and the collision handling is efficient.
Input array containing duplicate numbers that form valid tuples	The solution should correctly count valid tuple combinations even with duplicate numbers, without reusing the same index.
The array contains negative numbers	The product map should consider both positive and negative products.

Empty input array

How to Handle:

Return 0 immediately as there are no tuples to form.

Input array with fewer than 4 elements

How to Handle:

Return 0 as at least 4 elements are required to form two pairs.

Input array with very large numbers leading to potential integer overflow during multiplication

How to Handle:

Use a data type that can accommodate larger numbers or modulo the multiplication result to prevent overflow.

Input array containing zero

How to Handle:

Any product involving zero will always be zero; handle zero values separately to avoid invalid combinations.

Input array with all elements being 1

How to Handle:

Calculate the number of possible pairs directly using combinatorics.

Large input array with many pairs having the same product

How to Handle:

Ensure the HashMap storing products has sufficient capacity and the collision handling is efficient.

Input array containing duplicate numbers that form valid tuples

How to Handle:

The solution should correctly count valid tuple combinations even with duplicate numbers, without reusing the same index.

The array contains negative numbers

How to Handle:

The product map should consider both positive and negative products.