Problem of The Day: Tuple with Same Product

Problem Statement

Brute Force [TLE]

class Solution:
    def tupleSameProduct(self, nums: List[int]) -> int:
        self.res = 0
        N = len(nums)
        def dfs(idx, curr):
            if len(curr) == 4:
                if (curr[0] * curr[1]) == (curr[2] * curr[3]):
                    self.res += 1
                return
            if idx == N:
                return
            for i in range(N):
                nums[0], nums[i] = nums[i], nums[0]
                if nums[0] not in curr:
                    dfs(idx + 1, curr + [nums[0]])
                nums[0], nums[i] = nums[i], nums[0]


        dfs(0, [])
        return self.res

Understanding the Formula: ( \frac{x(x - 1)}{2} )

The formula:

[ \frac{x(x - 1)}{2} ]

is commonly used to calculate the number of unique pairs that can be formed from (x) items. This appears frequently in problems related to combinations, such as the handshake problem, network connections, or any scenario where you’re counting distinct pairs.

How to Derive the Formula

1. Identify the Problem

Imagine you have (x) people in a room, and you want to determine how many unique handshakes can occur if each person shakes hands with every other person exactly once.

2. Brute Force Counting (Example with 5 People)

Let’s manually count for 5 people (A, B, C, D, E):

• A shakes hands with B, C, D, E → 4 handshakes
• B has already shaken hands with A, so new handshakes with C, D, E → 3 handshakes
• C has already shaken hands with A and B, so new handshakes with D, E → 2 handshakes
• D has already shaken hands with A, B, C, so new handshake with E → 1 handshake
• E has already shaken hands with everyone → 0 new handshakes

Total handshakes:

[ 4 + 3 + 2 + 1 = 10 ]

3. Recognize the Pattern

Notice we are summing the first (4) positive integers:

[ (5 - 1) + (5 - 2) + (5 - 3) + (5 - 4) = 4 + 3 + 2 + 1 ]

For (x) people, the sum is:

[ (x - 1) + (x - 2) + \dots + 1 ]

This is the sum of the first (x - 1) positive integers.

4. Apply the Formula for the Sum of Integers

The formula for the sum of the first (n) positive integers is:

[ \text{Sum} = \frac{n(n + 1)}{2} ]

In our case, (n = x - 1). Therefore:

[ \text{Sum} = \frac{(x - 1) \cdot x}{2} ]

or equivalently:

[ \frac{x(x - 1)}{2} ]

5. Why Divide by 2?

When you form pairs, each connection is counted twice if you just multiply (x) by (x - 1):

• Person A shaking hands with B is the same as B shaking hands with A.

To avoid double-counting, we divide by 2.

Generalization

• Each item pairs with ((x - 1)) others.
• Avoid double-counting by dividing by 2.

Final Formula:

[ \frac{x(x - 1)}{2} ]

This formula is fundamental in combinatorics and helps solve problems involving unique pairings quickly and efficiently.

Editorial

Approach 2: Count Product Frequency

class Solution:
    def tupleSameProduct(self, nums):
        nums_length = len(nums)

        pair_products = []

        total_number_of_tuples = 0

        # Iterate over nums to calculate all pairwise products
        for first_index in range(nums_length):
            for second_index in range(first_index + 1, nums_length):
                pair_products.append(nums[first_index] * nums[second_index])

        pair_products.sort()

        last_product_seen = -1
        same_product_count = 0

        # Iterate over pair_products to count how many times each product occurs
        for product_index in range(len(pair_products)):
            if pair_products[product_index] == last_product_seen:
                # Increment the count of same products
                same_product_count += 1
            else:
                # Calculate how many pairs had the previous product value
                pairs_of_equal_product = (
                    (same_product_count - 1) * same_product_count // 2
                )

                total_number_of_tuples += 8 * pairs_of_equal_product

                # Update last_product_seen and reset same_product_count
                last_product_seen = pair_products[product_index]
                same_product_count = 1

        # Handle the last group of products (since the loop ends without adding
        # it)
        pairs_of_equal_product = (
            (same_product_count - 1) * same_product_count // 2
        )
        total_number_of_tuples += 8 * pairs_of_equal_product

        return total_number_of_tuples

Approach 3: Product Frequency Hash Map

class Solution(object):
    def tupleSameProduct(self, nums):
        nums_length = len(nums)

        # Initialize a dictionary to store the frequency of each product of pairs
        pair_products_frequency = {}

        total_number_of_tuples = 0

        # Iterate through each pair of numbers in `nums`
        for first_index in range(nums_length):
            for second_index in range(first_index + 1, nums_length):
                # Increment the frequency of the product of the current pair
                product_value = nums[first_index] * nums[second_index]
                if product_value in pair_products_frequency:
                    pair_products_frequency[product_value] += 1
                else:
                    pair_products_frequency[product_value] = 1

        # Iterate through each product value and its frequency in the dictionary
        for product_frequency in pair_products_frequency.values():
            # Calculate the number of ways to choose two pairs with the same product
            pairs_of_equal_product = (
                (product_frequency - 1) * product_frequency // 2
            )

            # Add the number of tuples for this product to the total (each pair
            # can form 8 tuples)
            total_number_of_tuples += 8 * pairs_of_equal_product

        return total_number_of_tuples