Minimum Absolute Sum Difference
To solve this problem, we follow these steps:
 Calculate Initial Absolute Sum Difference: Compute the initial absolute sum difference and find the difference for each element.
 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 innums2
.  Apply Best Replacement: Subtract the best difference found from the initial absolute sum difference.
 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:


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]
insorted_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 is3
, following the explanation given in the problem statement.
Alternative Approach


The Problem: You have two lists of numbers (
nums1
andnums2
), and you want to minimize the total absolute difference between corresponding elements of the two lists.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 innums1
to reduce the total difference.Finding the Best Replacement: For each position
i
, you want to find a number innums1
that, if replaced withnums1[i]
, would make the difference withnums2[i]
as small as possible.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.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 tonums2[i]
quickly.Applying Binary Search: You use a binary search on the sorted list to find the number in
nums1
that is closest tonums2[i]
. This is efficient and helps you find the best replacement quickly.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 timesaving.