Problem Description

In this problem, you are provided with two arrays of positive integers nums1 and nums2, both having the same length n. You need to calculate the absolute sum difference between the two arrays, which is the sum of the absolute differences between each pair of elements |nums1[i] - nums2[i]| at each index i ranging from 0 to n-1.

However, there's a twist. You have the option to replace one element from nums1 with any other element from the same nums1 array to minimize the overall absolute sum difference. After performing this optional replacement, you should return the minimum possible absolute sum difference. To account for potentially large numbers, the result should be presented modulo 10^9 + 7.

The goal is to find out how one can smartly replace a single element in nums1 to achieve the smallest possible absolute sum difference between nums1 and nums2.

Intuition

To approach this problem, one natural thought could be to check every possible single replacement in nums1 to minimize the absolute sum difference — but this would be inefficient for large arrays. To avoid an overly complicated and slow solution, we can use binary search, which is much more efficient.

The intuition here is based on finding, for each element nums2[i], the closest element in the sorted nums1. Since a sorted list is provided, binary search can be used to find this closest element efficiently.

Here's a step-by-step on how we achieve this:

1. Calculate the initial absolute sum difference without any replacements.
2. Sort a copy of nums1 to prepare for binary search.
3. Iterate through each element in nums2, and for each, search for the closest element in the sorted nums1 using binary search.
4. Determine the absolute difference reduction that could be achieved by replacing nums1[i] with this closest number.
5. Record the maximum amount by which we could reduce a single absolute difference.
6. Subtract this maximum reduction from the total absolute sum difference calculated in step 1 to achieve the minimum absolute sum difference after the single replacement.

Using binary search optimizes this process, because scanning through the sorted array for the nearest number would otherwise be a linear operation, which, when nested in another linear operation, would result in quadratic time complexity. Binary search turns the inner linear scan into a logarithmic one, thus significantly enhancing performance.

Learn more about Binary Search and Sorting patterns.

Solution Approach

The solution employs binary search, sorting, and modular arithmetic as its primary techniques:

• Sorting: A sorted version of nums1 is created, which allows for the efficient use of binary search. Sorting places elements in a sequence that binary search can exploit to reduce the time complexity from O(n) to O(log n) for searching.

• Binary Search: For each element b in nums2, we use binary search to find the position in the sorted nums1 where b could be inserted while maintaining the sort order. This gives us the element just larger than or equal to b as well as the one just smaller than b if one exists. These are the potential candidates for minimizing the difference with b if we were to make a replacement in nums1.

• Modular Arithmetic: Due to the constraints regarding large numbers, every sum is computed modulo 10^9 + 7, which is a prime number often used to avoid overflow issues in competitive programming.

Here's how the implementation reflects the algorithm:

1. Initialize mod = 10**9 + 7 to hold the value for modular arithmetic.
2. Calculate the initial sum of absolute differences s and mod it with mod.
3. Store the maximum reduction found as mx. Initially, mx is 0.
4. Iterate over the elements of nums1 and nums2.
   • For each pair (a, b), calculate the current difference d1.
   • Perform a binary search in the sorted nums to locate the insertion point for b.
   • If the insertion point is not at the beginning, calculate a potential reduced difference with the element just before the insertion point and update d2 if it's smaller.
   • If the insertion point is not at the end, also calculate the potential reduced difference with the element at the insertion point and update d2 if it's smaller.
   • Update mx to be the maximum difference that can be reduced for any element.
5. Finally, subtract this maximum achievable difference reduction mx from the initial sum s, and apply modular arithmetic to obtain the final result.

class Solution:
    def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
        mod = 10**9 + 7
        # Create a sorted copy of nums1 for [binary search](/problems/binary-search-speedrun).
        nums = sorted(nums1)
        # Calculate the initial sum of absolute differences.
        s = sum(abs(a - b) for a, b in zip(nums1, nums2)) % mod
        # Initialize the variable to track the maximum reduction possible.
        mx = 0
        # Iterate over each element of nums1 and nums2.
        for a, b in zip(nums1, nums2):
            # Calculate the current difference `d1`.
            d1 = abs(a - b)
            # Using binary search, find the position `i` to insert `b` in sorted `nums1`.
            i = bisect_left(nums, b)
            d2 = float('inf')  # Initialize a variable for the potential new difference.
            # Check the element at the insertion index if not at the end of the array.
            if i < len(nums):
                d2 = abs(nums[i] - b)
            # Check the element before the insertion index if possible to find a better candidate.
            if i:
                d2 = min(d2, abs(nums[i - 1] - b))
            # Update `mx` if we found a better reduction.
            mx = max(mx, d1 - d2)
        # The result is the initial sum `s` reduced by `mx` and then modded by `mod`.
        return (s - mx) % mod

By applying these techniques, the provided solution ensures that we find the minimum absolute sum difference in an optimized manner and handle potential overflow issues related to large sums effectively.

Example Walkthrough

Let's take the following small example to illustrate the solution approach:

nums1 = [1, 7, 5]
nums2 = [3, 9, 2]

We are to compute the absolute sum difference between these two arrays, while minimizing this value by replacing up to one element in nums1.

Step 1: Calculate the Initial Absolute Sum Difference

We first calculate the sum of absolute differences without any replacements:

|1 - 3| + |7 - 9| + |5 - 2| = 2 + 2 + 3 = 7

Step 2: Sort a Copy of nums1 for Binary Search

We then sort nums1 to:

nums = [1, 5, 7]

Step 3: Iterate and Binary Search for Each Element in nums2

Next, for each element in nums2, we look for the closest element in the sorted nums:

• For the first element in nums2 which is 3, the closest in nums is 1 (binary search would yield index 1 since 3 can be inserted before 5).
• For the second element 9, the closest is 7 (insertion index would be at the end of nums).
• For the third element 2, 1 is the closest in nums (binary search gives index 1 because 2 can be inserted before 5).

Step 4: Determine Absolute Difference Reduction

We now find the absolute difference reductions for each element in nums2:

• Reduction with 3 could be |1 - 3| = 2 to |5 - 3| = 2 (no reduction).
• Reduction with 9 could be |7 - 9| = 2 to |7 - 9| = 2 (no reduction).
• The most significant potential reduction comes from the element 2, where |5 - 2| = 3 can be reduced to |1 - 2| = 1. Here, a reduction of 2 is possible.

Step 5: Record the Maximum Absolute Difference Reduction

In this case, the reduction is 2 for the last element.

Step 6: Apply the Maximum Reduction to the Initial Sum

The initial sum was 7. After reducing the maximum reducible difference (2), we get 7 - 2 = 5.

Therefore, the minimum possible absolute sum difference after performing the optional replacement is 5, with no need for modular arithmetic since our numbers are small. In practice, we would take the result modulo 10^9 + 7 to ensure it fits within integer limits for larger sums.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code consists of several components:

1. Sorting nums1 takes O(n log n) time.
2. The list comprehension calculating the sum of absolute differences runs in O(n).
3. The for loop with binary search (bisect_left) inside runs in O(n log n) time, because bisect_left runs in O(log n) and is called up to twice for each of the n elements in nums1.

Thus, the overall time complexity is O(n log n + n + n log n), which simplifies to O(n log n).

Space Complexity

The space complexity of the code depends on the additional space required by:

1. The sorted list nums, which takes O(n) space.
2. The variables mod, s, and mx, which together use a constant amount of space O(1).

Consequently, the overall space complexity of the code is O(n) (due to the space needed for the sorted list).

Learn more about how to find time and space complexity quickly using problem constraints.