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Minimize Product Sum of Two Arrays

Medium

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Topics:

ArraysGreedy Algorithms

The product sum of two equal-length arrays a and b is equal to the sum of a[i] * b[i] for all 0 <= i < a.length (0-indexed).

For example, if a = [1,2,3,4] and b = [5,2,3,1], the product sum would be 1*5 + 2*2 + 3*3 + 4*1 = 22.

Given two arrays nums1 and nums2 of length n, return the minimum product sum if you are allowed to rearrange the nums1 array.

Example 1:

Input: nums1 = [5,3,4,2], nums2 = [4,2,2,5]
Output: 40
Explanation: We can rearrange nums1 to become [3,5,4,2]. The product sum of [3,5,4,2] and [4,2,2,5] is 3*4 + 5*2 + 4*2 + 2*5 = 40.

Example 2:

Input: nums1 = [2,1,4,5,7], nums2 = [3,2,4,8,6]
Output: 65
Explanation: We can rearrange nums1 to become [5,7,4,1,2]. The product sum of [5,7,4,1,2] and [3,2,4,8,6] is 5*3 + 7*2 + 4*4 + 1*8 + 2*6 = 65.

Constraints:

n == nums1.length == nums2.length
1 <= n <= 10⁵
-100 <= nums1[i], nums2[i] <= 100

Try coding it in LeetCode

Minimize Product Sum of Two Arrays

Medium

Asked by:

5 views

Topics:

ArraysGreedy Algorithms

The product sum of two equal-length arrays a and b is equal to the sum of a[i] * b[i] for all 0 <= i < a.length (0-indexed).

For example, if a = [1,2,3,4] and b = [5,2,3,1], the product sum would be 1*5 + 2*2 + 3*3 + 4*1 = 22.

Given two arrays nums1 and nums2 of length n, return the minimum product sum if you are allowed to rearrange the nums1 array.

Example 1:

Input: nums1 = [5,3,4,2], nums2 = [4,2,2,5]
Output: 40
Explanation: We can rearrange nums1 to become [3,5,4,2]. The product sum of [3,5,4,2] and [4,2,2,5] is 3*4 + 5*2 + 4*2 + 2*5 = 40.

Example 2:

Input: nums1 = [2,1,4,5,7], nums2 = [3,2,4,8,6]
Output: 65
Explanation: We can rearrange nums1 to become [5,7,4,1,2]. The product sum of [5,7,4,1,2] and [3,2,4,8,6] is 5*3 + 7*2 + 4*4 + 1*8 + 2*6 = 65.

Constraints:

n == nums1.length == nums2.length
1 <= n <= 10⁵
-100 <= nums1[i], nums2[i] <= 100

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 are the possible ranges and data types of the numbers within the arrays? Can they be negative, zero, or floating-point numbers?
What are the constraints on the sizes of the input arrays `nums1` and `nums2`? Are they guaranteed to be of the same length, and is there a minimum or maximum size?
If `nums1` and `nums2` are empty or `null`, what should the function return?
Are duplicate values allowed within `nums1` and `nums2`, and if so, how should they be handled in the minimization process?
Are we guaranteed that there is only one possible minimum product sum, or could there be multiple combinations that result in the same minimum value?

Brute Force Solution

Approach

The goal is to find the smallest possible result by strategically pairing numbers from two lists. The brute force way means trying every single way you could possibly pair the numbers together to see which one gives you the absolute smallest result.

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

Consider every possible way to mix up the order of the numbers in the first list.
For each possible order of the first list, consider every possible way to mix up the order of the numbers in the second list.
For each combination of orderings of both lists, multiply the corresponding numbers in the two lists and add them all up.
Remember this sum.
Keep doing steps 2-4 for every single possible combination of orderings of the two lists.
Once you have calculated every possible sum, look at all the sums you remembered, and find the smallest one.

Code Implementation

import itertools

def minimize_product_sum_brute_force(first_list, second_list):
    minimum_product_sum = float('inf')

    # Iterate through all permutations of the first list
    for first_list_permutation in itertools.permutations(first_list):

        # Iterate through all permutations of the second list
        for second_list_permutation in itertools.permutations(second_list):
            current_product_sum = 0

            # Calculate the product sum for the current permutations
            for index in range(len(first_list)):
                current_product_sum += first_list_permutation[index] * second_list_permutation[index]

            # Track the smallest product sum
            minimum_product_sum = min(minimum_product_sum, current_product_sum)

    return minimum_product_sum

Big(O) Analysis

Time Complexity

O(n! * n!) – The algorithm considers every possible permutation of nums1 and nums2. Generating all permutations of an array of size n has a time complexity of O(n!). Since we need to generate all permutations of both nums1 and nums2, the cost becomes O(n! * n!). The multiplication and summation within each permutation pairing takes O(n) time, but that is dwarfed by the cost of permutation generation. Thus, the overall time complexity is dominated by the permutation generation, resulting in O(n! * n!).

Space Complexity

O(1) – The described brute force approach, which iterates through all possible permutations of both input arrays, does not explicitly use any auxiliary data structures besides a variable to store the current product sum and a variable to keep track of the minimum product sum. The permutations are generated implicitly through nested loops or recursion, potentially modifying the input arrays in place (although the explanation doesn't state this explicitly, we're analyzing only the space required based on the plain English explanation). Therefore, the auxiliary space required remains constant regardless of the input size N, where N is the number of elements in each array.

Optimal Solution

Approach

To minimize the product sum, we want to pair the largest number in one list with the smallest number in the other. This strategy avoids needlessly multiplying large numbers together. We achieve this by sorting one list in ascending order and the other in descending order, then calculating the sum of the products of corresponding elements.

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

Arrange the first list of numbers from smallest to largest.
Arrange the second list of numbers from largest to smallest.
Multiply the first number from the first list by the first number from the second list.
Repeat the previous step for each corresponding pair of numbers from both lists.
Add up all the products from the previous steps to get the smallest possible total.

Code Implementation

def minimize_product_sum(first_numbers, second_numbers):
    # Sort the first list in ascending order.
    first_numbers.sort()

    # Sort the second list in descending order.
    second_numbers.sort(reverse=True)

    minimum_product_sum = 0

    for index in range(len(first_numbers)):
        # Calculate the product sum.
        minimum_product_sum += first_numbers[index] * second_numbers[index]

    return minimum_product_sum

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation is sorting the two input arrays, nums1 and nums2, each containing n elements. Common sorting algorithms, such as merge sort or quicksort, have a time complexity of O(n log n). Since we perform this sorting operation twice (once for each array), the total time complexity is O(n log n) + O(n log n), which simplifies to O(n log n). The subsequent multiplication and summation of corresponding elements takes O(n) time, but this is less significant than the sorting cost, so it doesn't affect the overall Big O complexity.

Space Complexity

O(N) – The provided solution involves sorting both input lists. Most sorting algorithms, such as merge sort or quicksort, used in standard library sort functions, require auxiliary space. These algorithms often create temporary arrays to hold the sorted elements during the sorting process. In the worst-case scenario, the temporary arrays can be of size N, where N is the number of elements in the input lists. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
One or both input arrays are null	Throw IllegalArgumentException or return a specific error value like -1 to signify invalid input.
One or both input arrays are empty	Return 0 as the product sum will be zero or throw an exception indicating invalid input, depending on problem specifications.
Input arrays have different lengths	Throw IllegalArgumentException or return a specific error value as the product sum cannot be computed.
Arrays contain only positive numbers	The standard solution of sorting one array ascending and the other descending works correctly.
Arrays contain only negative numbers	The standard solution of sorting one array ascending and the other descending still provides the minimized sum.
Arrays contain mixed positive, negative, and zero values	Sorting one array ascending and the other descending remains the correct approach to minimize the product sum.
Arrays contain very large numbers that could lead to integer overflow when multiplied	Use long data type for calculations and consider using BigInteger for extremely large numbers to avoid overflow.
Extremely large arrays to test the performance of sorting algorithms	Ensure that a O(n log n) sorting algorithm is used to handle large arrays efficiently; consider using optimized implementations like merge sort or quicksort.

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 are the possible ranges and data types of the numbers within the arrays? Can they be negative, zero, or floating-point numbers?
What are the constraints on the sizes of the input arrays `nums1` and `nums2`? Are they guaranteed to be of the same length, and is there a minimum or maximum size?
If `nums1` and `nums2` are empty or `null`, what should the function return?
Are duplicate values allowed within `nums1` and `nums2`, and if so, how should they be handled in the minimization process?
Are we guaranteed that there is only one possible minimum product sum, or could there be multiple combinations that result in the same minimum value?

Brute Force Solution

Approach

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

Consider every possible way to mix up the order of the numbers in the first list.
For each possible order of the first list, consider every possible way to mix up the order of the numbers in the second list.
For each combination of orderings of both lists, multiply the corresponding numbers in the two lists and add them all up.
Remember this sum.
Keep doing steps 2-4 for every single possible combination of orderings of the two lists.
Once you have calculated every possible sum, look at all the sums you remembered, and find the smallest one.

Code Implementation

import itertools

def minimize_product_sum_brute_force(first_list, second_list):
    minimum_product_sum = float('inf')

    # Iterate through all permutations of the first list
    for first_list_permutation in itertools.permutations(first_list):

        # Iterate through all permutations of the second list
        for second_list_permutation in itertools.permutations(second_list):
            current_product_sum = 0

            # Calculate the product sum for the current permutations
            for index in range(len(first_list)):
                current_product_sum += first_list_permutation[index] * second_list_permutation[index]

            # Track the smallest product sum
            minimum_product_sum = min(minimum_product_sum, current_product_sum)

    return minimum_product_sum

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Arrange the first list of numbers from smallest to largest.
Arrange the second list of numbers from largest to smallest.
Multiply the first number from the first list by the first number from the second list.
Repeat the previous step for each corresponding pair of numbers from both lists.
Add up all the products from the previous steps to get the smallest possible total.

Code Implementation

def minimize_product_sum(first_numbers, second_numbers):
    # Sort the first list in ascending order.
    first_numbers.sort()

    # Sort the second list in descending order.
    second_numbers.sort(reverse=True)

    minimum_product_sum = 0

    for index in range(len(first_numbers)):
        # Calculate the product sum.
        minimum_product_sum += first_numbers[index] * second_numbers[index]

    return minimum_product_sum

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
One or both input arrays are null	Throw IllegalArgumentException or return a specific error value like -1 to signify invalid input.
One or both input arrays are empty	Return 0 as the product sum will be zero or throw an exception indicating invalid input, depending on problem specifications.
Input arrays have different lengths	Throw IllegalArgumentException or return a specific error value as the product sum cannot be computed.
Arrays contain only positive numbers	The standard solution of sorting one array ascending and the other descending works correctly.
Arrays contain only negative numbers	The standard solution of sorting one array ascending and the other descending still provides the minimized sum.
Arrays contain mixed positive, negative, and zero values	Sorting one array ascending and the other descending remains the correct approach to minimize the product sum.
Arrays contain very large numbers that could lead to integer overflow when multiplied	Use long data type for calculations and consider using BigInteger for extremely large numbers to avoid overflow.
Extremely large arrays to test the performance of sorting algorithms	Ensure that a O(n log n) sorting algorithm is used to handle large arrays efficiently; consider using optimized implementations like merge sort or quicksort.