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Maximum Product Difference Between Two Pairs

#568 Most AskedEasy

Maximum Product Difference Between Two Pairs #568 Most Asked

Easy

Google

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Topics:

ArraysGreedy Algorithms

The product difference between two pairs (a, b) and (c, d) is defined as (a * b) - (c * d).

For example, the product difference between (5, 6) and (2, 7) is (5 * 6) - (2 * 7) = 16.

Given an integer array nums, choose four distinct indices w, x, y, and z such that the product difference between pairs (nums[w], nums[x]) and (nums[y], nums[z]) is maximized.

Return the maximum such product difference.

Example 1:

Input: nums = [5,6,2,7,4]
Output: 34
Explanation: We can choose indices 1 and 3 for the first pair (6, 7) and indices 2 and 4 for the second pair (2, 4).
The product difference is (6 * 7) - (2 * 4) = 34.

Example 2:

Input: nums = [4,2,5,9,7,4,8]
Output: 64
Explanation: We can choose indices 3 and 6 for the first pair (9, 8) and indices 1 and 5 for the second pair (2, 4).
The product difference is (9 * 8) - (2 * 4) = 64.

Constraints:

4 <= nums.length <= 10⁴
1 <= nums[i] <= 10⁴

Try coding it in LeetCode

Maximum Product Difference Between Two Pairs

#568 Most AskedEasy

Maximum Product Difference Between Two Pairs #568 Most Asked

Easy

Google

8 views

Topics:

ArraysGreedy Algorithms

The product difference between two pairs (a, b) and (c, d) is defined as (a * b) - (c * d).

For example, the product difference between (5, 6) and (2, 7) is (5 * 6) - (2 * 7) = 16.

Return the maximum such product difference.

Example 1:

Input: nums = [5,6,2,7,4]
Output: 34
Explanation: We can choose indices 1 and 3 for the first pair (6, 7) and indices 2 and 4 for the second pair (2, 4).
The product difference is (6 * 7) - (2 * 4) = 34.

Example 2:

Input: nums = [4,2,5,9,7,4,8]
Output: 64
Explanation: We can choose indices 3 and 6 for the first pair (9, 8) and indices 1 and 5 for the second pair (2, 4).
The product difference is (9 * 8) - (2 * 4) = 64.

Constraints:

4 <= nums.length <= 10⁴
1 <= nums[i] <= 10⁴

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 range of values for the numbers within the input array? Can the array contain negative numbers, zero, or only positive numbers?
What is the minimum size of the array? Is an empty or single-element array a possible input?
Can the input array contain duplicate values, and if so, how should duplicates be handled when finding the maximum product difference?
Is the input array guaranteed to have at least four elements?
If multiple pairs result in the same maximum product difference, is any one of them acceptable?

Brute Force Solution

Approach

The goal is to find two pairs of numbers from a given set, and maximize the difference between their products. The brute force method involves checking every single possible pair of pairs to find the one that yields the largest product difference.

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

First, consider every possible combination of four numbers from the given set.
For each combination of four numbers, form two pairs in all possible ways.
For each pair of pairs, calculate the product of each pair and then find the difference between these two products.
Keep track of the largest difference you've found so far.
After checking all combinations of four numbers and all ways to make pairs from each combination, the largest difference you kept track of is the answer.

Code Implementation

def max_product_difference_brute_force(numbers):
    max_difference = 0
    list_length = len(numbers)

    # Iterate through all possible combinations of four numbers.
    for first_index in range(list_length):
        for second_index in range(first_index + 1, list_length):
            for third_index in range(second_index + 1, list_length):
                for fourth_index in range(third_index + 1, list_length):
                    first_number = numbers[first_index]
                    second_number = numbers[second_index]
                    third_number = numbers[third_index]
                    fourth_number = numbers[fourth_index]

                    # Create all possible pairs from the four numbers
                    product_one = first_number * second_number
                    product_two = third_number * fourth_number
                    difference = abs(product_one - product_two)
                    max_difference = max(max_difference, difference)

                    product_one = first_number * third_number
                    product_two = second_number * fourth_number
                    difference = abs(product_one - product_two)
                    max_difference = max(max_difference, difference)

                    product_one = first_number * fourth_number
                    product_two = second_number * third_number
                    difference = abs(product_one - product_two)

                    # Keep track of the maximum difference found so far.
                    max_difference = max(max_difference, difference)

    return max_difference

Big(O) Analysis

Time Complexity

O(n^4) – The provided solution involves selecting every possible combination of four numbers from the input array of size n. Choosing four elements has a complexity of O(n^4). For each of these combinations, the algorithm then considers all possible pairings of the four selected numbers. The number of ways to pair four numbers is constant (limited to a few possibilities). After pairing, we perform constant time operations to compute the product difference. Since the dominant operation is choosing combinations of four elements which is O(n^4), the overall time complexity is O(n^4).

Space Complexity

O(1) – The provided brute force solution iterates through combinations and pairs but does not explicitly create auxiliary data structures that scale with the input size N (the size of the input set). It keeps track of the largest difference found so far using a single variable. Therefore, the extra space used is constant regardless of the input size N, resulting in O(1) space complexity.

Optimal Solution

Approach

To find the biggest product difference quickly, we don't need to check every possible pair. We just need to identify the two largest and two smallest numbers, and then calculate the difference between their products.

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

Find the two biggest numbers from the group.
Find the two smallest numbers from the group.
Multiply the two biggest numbers together.
Multiply the two smallest numbers together.
Subtract the smaller product from the bigger product. This is the biggest possible product difference.

Code Implementation

def maximum_product_difference(numbers):
    numbers.sort()

    # Finding two smallest numbers.
    smallest_number_one = numbers[0]

    smallest_number_two = numbers[1]

    # Finding two largest numbers.
    largest_number_one = numbers[-1]

    largest_number_two = numbers[-2]

    product_of_largest_numbers = largest_number_one * largest_number_two

    product_of_smallest_numbers = smallest_number_one * smallest_number_two

    # Calculate the difference as required.
    maximum_product_difference_value = product_of_largest_numbers - product_of_smallest_numbers

    return maximum_product_difference_value

Big(O) Analysis

Time Complexity

O(n) – The described algorithm requires finding the two largest and two smallest numbers in the input array. This can be done by iterating through the array once, keeping track of the current largest, second largest, smallest, and second smallest numbers. Each element in the array of size n is visited a constant number of times (to potentially update the min/max values). Therefore, the time complexity is directly proportional to the input size n, resulting in a linear time complexity of O(n).

Space Complexity

O(1) – The algorithm identifies the two largest and two smallest numbers. This can be done by maintaining a constant number of variables to store these four values, regardless of the input array's size (N). No auxiliary data structures that scale with the input are used. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 immediately, as no product difference can be calculated.
Array with fewer than 4 elements	Return 0, because a minimum of four elements is required to calculate the product difference between two pairs.
Array with all elements being identical	The largest and smallest two elements will all be the same, resulting in a product difference of 0, which is the correct result.
Array with a large number of elements (performance)	Using an efficient O(n) algorithm like finding the four extreme values should scale linearly and avoid timeouts.
Array containing very large positive integers (potential overflow)	Use a data type that can accommodate large numbers (e.g., long in Java/C++) to prevent integer overflow during multiplication.
Array containing only negative numbers	The two largest numbers (least negative) should be selected for the smallest pair and two smallest (most negative) for the largest pair.
Array containing a mix of positive, negative, and zero values	The algorithm should correctly identify the two largest positive and two smallest negative (or zero) numbers for calculation.
Array contains duplicate minimum or maximum values.	The algorithm must correctly identify duplicates if they contribute to one of the smallest or largest pairs by either not considering same index or finding all indexes and choosing different ones.

Empty or null input array

How to Handle:

Return 0 immediately, as no product difference can be calculated.

Array with fewer than 4 elements

How to Handle:

Return 0, because a minimum of four elements is required to calculate the product difference between two pairs.

Array with all elements being identical

How to Handle:

The largest and smallest two elements will all be the same, resulting in a product difference of 0, which is the correct result.

Array with a large number of elements (performance)

How to Handle:

Using an efficient O(n) algorithm like finding the four extreme values should scale linearly and avoid timeouts.

Array containing very large positive integers (potential overflow)

How to Handle:

Use a data type that can accommodate large numbers (e.g., long in Java/C++) to prevent integer overflow during multiplication.

Array containing only negative numbers

How to Handle:

The two largest numbers (least negative) should be selected for the smallest pair and two smallest (most negative) for the largest pair.

Array containing a mix of positive, negative, and zero values

How to Handle:

The algorithm should correctly identify the two largest positive and two smallest negative (or zero) numbers for calculation.

Array contains duplicate minimum or maximum values.

How to Handle:

The algorithm must correctly identify duplicates if they contribute to one of the smallest or largest pairs by either not considering same index or finding all indexes and choosing different ones.

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 range of values for the numbers within the input array? Can the array contain negative numbers, zero, or only positive numbers?
What is the minimum size of the array? Is an empty or single-element array a possible input?
Can the input array contain duplicate values, and if so, how should duplicates be handled when finding the maximum product difference?
Is the input array guaranteed to have at least four elements?
If multiple pairs result in the same maximum product difference, is any one of them acceptable?

Brute Force Solution

Approach

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

First, consider every possible combination of four numbers from the given set.
For each combination of four numbers, form two pairs in all possible ways.
For each pair of pairs, calculate the product of each pair and then find the difference between these two products.
Keep track of the largest difference you've found so far.
After checking all combinations of four numbers and all ways to make pairs from each combination, the largest difference you kept track of is the answer.

Code Implementation

def max_product_difference_brute_force(numbers):
    max_difference = 0
    list_length = len(numbers)

    # Iterate through all possible combinations of four numbers.
    for first_index in range(list_length):
        for second_index in range(first_index + 1, list_length):
            for third_index in range(second_index + 1, list_length):
                for fourth_index in range(third_index + 1, list_length):
                    first_number = numbers[first_index]
                    second_number = numbers[second_index]
                    third_number = numbers[third_index]
                    fourth_number = numbers[fourth_index]

                    # Create all possible pairs from the four numbers
                    product_one = first_number * second_number
                    product_two = third_number * fourth_number
                    difference = abs(product_one - product_two)
                    max_difference = max(max_difference, difference)

                    product_one = first_number * third_number
                    product_two = second_number * fourth_number
                    difference = abs(product_one - product_two)
                    max_difference = max(max_difference, difference)

                    product_one = first_number * fourth_number
                    product_two = second_number * third_number
                    difference = abs(product_one - product_two)

                    # Keep track of the maximum difference found so far.
                    max_difference = max(max_difference, difference)

    return max_difference

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Find the two biggest numbers from the group.
Find the two smallest numbers from the group.
Multiply the two biggest numbers together.
Multiply the two smallest numbers together.
Subtract the smaller product from the bigger product. This is the biggest possible product difference.

Code Implementation

def maximum_product_difference(numbers):
    numbers.sort()

    # Finding two smallest numbers.
    smallest_number_one = numbers[0]

    smallest_number_two = numbers[1]

    # Finding two largest numbers.
    largest_number_one = numbers[-1]

    largest_number_two = numbers[-2]

    product_of_largest_numbers = largest_number_one * largest_number_two

    product_of_smallest_numbers = smallest_number_one * smallest_number_two

    # Calculate the difference as required.
    maximum_product_difference_value = product_of_largest_numbers - product_of_smallest_numbers

    return maximum_product_difference_value

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 immediately, as no product difference can be calculated.
Array with fewer than 4 elements	Return 0, because a minimum of four elements is required to calculate the product difference between two pairs.
Array with all elements being identical	The largest and smallest two elements will all be the same, resulting in a product difference of 0, which is the correct result.
Array with a large number of elements (performance)	Using an efficient O(n) algorithm like finding the four extreme values should scale linearly and avoid timeouts.
Array containing very large positive integers (potential overflow)	Use a data type that can accommodate large numbers (e.g., long in Java/C++) to prevent integer overflow during multiplication.
Array containing only negative numbers	The two largest numbers (least negative) should be selected for the smallest pair and two smallest (most negative) for the largest pair.
Array containing a mix of positive, negative, and zero values	The algorithm should correctly identify the two largest positive and two smallest negative (or zero) numbers for calculation.
Array contains duplicate minimum or maximum values.	The algorithm must correctly identify duplicates if they contribute to one of the smallest or largest pairs by either not considering same index or finding all indexes and choosing different ones.

Empty or null input array

How to Handle:

Return 0 immediately, as no product difference can be calculated.

Array with fewer than 4 elements

How to Handle:

Return 0, because a minimum of four elements is required to calculate the product difference between two pairs.

Array with all elements being identical

How to Handle:

The largest and smallest two elements will all be the same, resulting in a product difference of 0, which is the correct result.

Array with a large number of elements (performance)

How to Handle:

Using an efficient O(n) algorithm like finding the four extreme values should scale linearly and avoid timeouts.

Array containing very large positive integers (potential overflow)

How to Handle:

Use a data type that can accommodate large numbers (e.g., long in Java/C++) to prevent integer overflow during multiplication.

Array containing only negative numbers

How to Handle:

The two largest numbers (least negative) should be selected for the smallest pair and two smallest (most negative) for the largest pair.

Array containing a mix of positive, negative, and zero values

How to Handle:

The algorithm should correctly identify the two largest positive and two smallest negative (or zero) numbers for calculation.

Array contains duplicate minimum or maximum values.

How to Handle:

The algorithm must correctly identify duplicates if they contribute to one of the smallest or largest pairs by either not considering same index or finding all indexes and choosing different ones.

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