Minimum Absolute Sum Difference

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ArraysBinary Search

You are given two positive integer arrays nums1 and nums2, both of length n.

The absolute sum difference of arrays nums1 and nums2 is defined as the sum of |nums1[i] - nums2[i]| for each 0 <= i < n (0-indexed).

You can replace at most one element of nums1 with any other element in nums1 to minimize the absolute sum difference.

Return the minimum absolute sum difference after replacing at most one element in the array nums1. Since the answer may be large, return it modulo 10⁹ + 7.

|x| is defined as:

x if x >= 0, or
-x if x < 0.

Example 1:

Input: nums1 = [1,7,5], nums2 = [2,3,5]
Output: 3
Explanation: There are two possible optimal solutions:
- Replace the second element with the first: [1,7,5] => [1,1,5], or
- Replace the second element with the third: [1,7,5] => [1,5,5].
Both will yield an absolute sum difference of |1-2| + (|1-3| or |5-3|) + |5-5| = 3.

Example 2:

Input: nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10]
Output: 0
Explanation: nums1 is equal to nums2 so no replacement is needed. This will result in an 
absolute sum difference of 0.

Example 3:

Input: nums1 = [1,10,4,4,2,7], nums2 = [9,3,5,1,7,4]
Output: 20
Explanation: Replace the first element with the second: [1,10,4,4,2,7] => [10,10,4,4,2,7].
This yields an absolute sum difference of |10-9| + |10-3| + |4-5| + |4-1| + |2-7| + |7-4| = 20

Constraints:

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

Try coding it in LeetCode

Minimum Absolute Sum Difference

Medium

Asked by:

7 views

Topics:

ArraysBinary Search

You are given two positive integer arrays nums1 and nums2, both of length n.

The absolute sum difference of arrays nums1 and nums2 is defined as the sum of |nums1[i] - nums2[i]| for each 0 <= i < n (0-indexed).

You can replace at most one element of nums1 with any other element in nums1 to minimize the absolute sum difference.

Return the minimum absolute sum difference after replacing at most one element in the array nums1. Since the answer may be large, return it modulo 10⁹ + 7.

|x| is defined as:

x if x >= 0, or
-x if x < 0.

Example 1:

Input: nums1 = [1,7,5], nums2 = [2,3,5]
Output: 3
Explanation: There are two possible optimal solutions:
- Replace the second element with the first: [1,7,5] => [1,1,5], or
- Replace the second element with the third: [1,7,5] => [1,5,5].
Both will yield an absolute sum difference of |1-2| + (|1-3| or |5-3|) + |5-5| = 3.

Example 2:

Input: nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10]
Output: 0
Explanation: nums1 is equal to nums2 so no replacement is needed. This will result in an 
absolute sum difference of 0.

Example 3:

Input: nums1 = [1,10,4,4,2,7], nums2 = [9,3,5,1,7,4]
Output: 20
Explanation: Replace the first element with the second: [1,10,4,4,2,7] => [10,10,4,4,2,7].
This yields an absolute sum difference of |10-9| + |10-3| + |4-5| + |4-1| + |2-7| + |7-4| = 20

Constraints:

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

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 for the numbers within the input arrays `nums1` and `nums2`? Specifically, can they be negative, zero, or floating-point numbers?
What are the size constraints for the input arrays `nums1` and `nums2`? Will they always be of the same length and can I assume they are non-empty?
If multiple modifications to `nums1` result in the same minimum absolute sum difference, is any one of them acceptable, or is there a specific tie-breaking condition?
Could you clarify the 'absolute sum difference'? Should I sum the absolute differences between elements at corresponding indices across the two arrays?
If all elements in `nums1` are already optimal (i.e., replacing them doesn't reduce the absolute sum difference), what value should be returned?

Brute Force Solution

Approach

The brute force approach to minimizing the absolute sum difference involves checking every possible replacement in the first set of numbers. We try replacing each number in the first set with every number from the second set to see which replacement minimizes the overall difference.

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

Start by calculating the initial total absolute difference between the two sets of numbers without any replacements.
Now, consider each number in the first set, one at a time.
For each number in the first set, try replacing it with every single number from the second set.
For each replacement, calculate the new total absolute difference between the modified first set and the second set.
Compare this new total difference with the smallest total difference found so far. If it's smaller, remember it and the replacement you made.
After trying all possible replacements for that number in the first set, move on to the next number in the first set.
Repeat this process until you've tried replacing every number in the first set with every number in the second set.
At the end, you'll have the smallest possible total absolute difference you found across all the replacements and the corresponding replacements that produced that difference.

Code Implementation

def min_absolute_sum_difference_brute_force(numbers1, numbers2):
    total_absolute_difference = 0
    for index in range(len(numbers1)):
        total_absolute_difference += abs(numbers1[index] - numbers2[index])

    minimum_total_difference = total_absolute_difference

    for index1 in range(len(numbers1)):
        for index2 in range(len(numbers2)):
            original_value = numbers1[index1]
            numbers1[index1] = numbers2[index2]
            
            # Recalculate the total absolute difference with the replacement
            new_total_absolute_difference = 0
            for index in range(len(numbers1)):
                new_total_absolute_difference += abs(numbers1[index] - numbers2[index])
            
            #Check if current difference is smaller than the minimum seen
            if new_total_absolute_difference < minimum_total_difference:
                minimum_total_difference = new_total_absolute_difference

            #Revert back to the original value for next iteration
            numbers1[index1] = original_value

    return minimum_total_difference

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n elements in the first array (nums1). For each of these elements, it then iterates through all n elements in the second array (nums2) to find the best replacement. This nested loop structure results in n * n comparisons. Thus, the total operations approximate to n*n, so the time complexity is O(n²).

Space Complexity

O(1) – The provided brute force algorithm calculates the minimum absolute sum difference by iteratively comparing the current total difference with potentially better ones after replacing elements. It keeps track of the smallest total difference found so far, which requires a single variable. The algorithm does not use any auxiliary data structures like arrays, hash maps, or recursion. Therefore, the auxiliary space complexity is constant, or O(1).

Optimal Solution

Approach

The key is to minimize the absolute difference by strategically replacing elements in one set of numbers. We want to find the element in the second set that's closest to each element in the first set, and see if replacing improves the total difference.

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

For each number in the first set, find the number in the second set that's closest to it.
Calculate the original difference between the two numbers from the first and second set.
Calculate the new difference you'd get if you replaced the first number with the closest number from the second set.
See if this new difference is smaller than the original difference. If it is, remember how much smaller it is.
Repeat this for all the numbers in the first set. Keep track of the biggest improvement you can make by replacing a number.
After checking all numbers, subtract the biggest possible improvement from the total original difference. Make sure the final result isn't negative.

Code Implementation

def min_absolute_sum_difference(first_array, second_array):
    total_absolute_difference = 0
    maximum_reduction = 0
    sorted_second_array = sorted(second_array)
    array_length = len(first_array)

    for index in range(array_length):
        first_array_element = first_array[index]
        original_difference = abs(first_array_element - second_array[index])
        total_absolute_difference += original_difference

        # Find the closest element in the second array using binary search.
        left_pointer = 0
        right_pointer = array_length - 1
        closest_element = None

        while left_pointer <= right_pointer:
            middle_pointer = (left_pointer + right_pointer) // 2
            potential_closest = sorted_second_array[middle_pointer]

            if closest_element is None or abs(first_array_element - potential_closest) < abs(first_array_element - closest_element):
                closest_element = potential_closest

            if potential_closest < first_array_element:
                left_pointer = middle_pointer + 1
            else:
                right_pointer = middle_pointer - 1

        # Calculate the potential reduction by replacing.
        if closest_element is not None:
            new_difference = abs(first_array_element - closest_element)
            reduction = original_difference - new_difference

            # Keep track of the maximum possible reduction.
            if reduction > maximum_reduction:
                maximum_reduction = reduction

    # Subtract the maximum possible reduction from the total difference.
    result = (total_absolute_difference - maximum_reduction) % (10**9 + 7)
    return result

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm iterates through each of the n elements of the first array. For each element, it finds the closest element in the second array. Finding the closest element in a sorted array of size n can be done in O(log n) time using binary search. Therefore, the dominant operation is performing the binary search n times, leading to a time complexity of O(n log n).

Space Complexity

O(N) – The algorithm's space complexity stems from sorting the second set of numbers. Sorting algorithms like merge sort or quicksort, commonly used by built-in sort functions, may require auxiliary space proportional to the number of elements being sorted. The plain English explanation doesn't specify which sorting algorithm is used but implies that sorting is needed for finding the closest number in the second set. Therefore, an auxiliary array of size N might be needed during sorting, where N is the number of elements in the second set. This makes the overall auxiliary space complexity O(N).

Edge Cases

Case	How to Handle
Empty nums1 or nums2	Return 0, as the absolute sum difference is 0 for empty arrays.
nums1 and nums2 have different lengths	The problem statement should specify if the lengths are different, or if the function must throw an error if they are.
nums1 and nums2 contain extremely large positive/negative integers that could cause overflow.	Use long data type to prevent integer overflow when calculating absolute differences and sums, and apply the modulo operation during accumulation.
All elements in nums1 are identical.	The binary search should handle this edge case gracefully and still find the best replacement in nums2.
nums2 contains only one element.	Binary search in nums2 simplifies to checking if replacing any element in nums1 with the single element in nums2 reduces the difference.
Large input size leading to potential time complexity issues.	Ensure binary search or similar efficient methods are used to find optimal replacements to achieve O(n log n) time complexity, rather than O(n^2).
Modulus operation can result in negative values (depending on the programming language).	Always ensure the result of the modulus operation is non-negative by adding the modulus value when the result is negative.
The initial absolute sum difference is already 0.	The algorithm should still execute correctly and potentially find other replacements that also result in an absolute sum difference of 0, although no further reduction is possible.