Minimum Absolute Sum Difference

To solve this problem, we follow these steps:

  1. Calculate Initial Absolute Sum Difference: Compute the initial absolute sum difference and find the difference for each element.
  2. Find Best Replacement: Iterate through the array and find the best possible replacement for each element in nums1 that would minimize the absolute difference with the corresponding element in nums2.
  3. Apply Best Replacement: Subtract the best difference found from the initial absolute sum difference.
  4. Return Result: The result is the minimum absolute sum difference after applying the best replacement. Since the answer may be large, return it modulo (10^9 + 7).

Here’s the code:

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from bisect import bisect_left

class Solution:
    def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
        MOD = 10**9 + 7
        n = len(nums1)

        # Sorted version of nums1 for binary search
        sorted_nums1 = sorted(nums1)

        # Calculate the initial absolute sum difference
        total_diff = 0
        diffs = []
        for i in range(n):
            diff = abs(nums1[i] - nums2[i])
            total_diff += diff
            diffs.append(diff)

        # Find the best replacement
        max_diff_improvement = 0
        for i in range(n):
            # Skip if no difference
            if diffs[i] == 0:
                continue

            # Use binary search to find the best replacement
            idx = bisect_left(sorted_nums1, nums2[i])
            if idx > 0:
                max_diff_improvement = max(max_diff_improvement, diffs[i] - abs(sorted_nums1[idx - 1] - nums2[i]))
            if idx < n:
                max_diff_improvement = max(max_diff_improvement, diffs[i] - abs(sorted_nums1[idx] - nums2[i]))

        # Apply best replacement and return the result modulo 10^9 + 7
        return (total_diff - max_diff_improvement) % MOD

Explanation:

  • We calculate the initial absolute sum difference and store the difference for each pair in diffs.
  • We sort nums1 to apply binary search for finding the best replacement.
  • We use binary search to find the closest number to nums2[i] in sorted_nums1 and calculate the improvement in the difference.
  • Finally, we return the result after applying the best replacement.

Example:

  • For the input nums1 = [1,7,5], nums2 = [2,3,5], the output is 3, following the explanation given in the problem statement.

Alternative Approach

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from typing import List

class Solution:
    def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
        res = 0
        gain = 0
        s = sorted(set(nums1))
        for i in range(len(nums1)):
            original = abs(nums1[i] - nums2[i])
            res += original
            if gain < original:
                idx = self.binary_search(s, nums2[i])
                if idx < len(s):
                    gain = max(gain, original - abs(s[idx] - nums2[i]))
                if idx > 0:
                    gain = max(gain, original - abs(s[idx - 1] - nums2[i]))
        return (res - gain) % 1000000007

    def binary_search(self, arr, target):
        left, right = 0, len(arr) - 1
        while left <= right:
            mid = left + (right - left) // 2
            if arr[mid] < target:
                left = mid + 1
            else:
                right = mid - 1
        return left

  1. The Problem: You have two lists of numbers (nums1 and nums2), and you want to minimize the total absolute difference between corresponding elements of the two lists.

  2. Understanding the Difference: For each pair of corresponding elements, the absolute difference is given by abs(nums1[i] - nums2[i]). You can change one element in nums1 to reduce the total difference.

  3. Finding the Best Replacement: For each position i, you want to find a number in nums1 that, if replaced with nums1[i], would make the difference with nums2[i] as small as possible.

  4. Tracking the Gain: You track the biggest gain that you can get by replacing one number in nums1. The gain is the reduction in the difference that a replacement would bring.

  5. Using Sorted List: By sorting nums1 and keeping it in a separate list (s), you can quickly search for the best number to use for replacement. Sorting allows you to find numbers that are close to nums2[i] quickly.

  6. Applying Binary Search: You use a binary search on the sorted list to find the number in nums1 that is closest to nums2[i]. This is efficient and helps you find the best replacement quickly.

  7. Final Calculation: You subtract the biggest gain from the total absolute difference to get the minimum absolute sum difference.

Metaphor

Think of nums1 and nums2 as two stacks of blocks. Each block in nums1 is paired with a block in nums2, and the height difference between each pair of blocks is what you’re trying to minimize.

You’re allowed to replace one block in nums1 with another block from a spare set (s) to reduce the overall height difference. By sorting the spare set, you can quickly find the best block to use for replacement.

In the end, you replace the block in nums1 that results in the largest reduction in height difference, thus minimizing the total difference between the two stacks of blocks.

This approach leverages sorting and binary search to find the best replacement efficiently, ensuring a solution that is both effective and time-saving.