Minimum Absolute Sum Difference - Leetcode Solution

Problem statement: Given two arrays of integers, nums1 and nums2, return the minimum absolute sum difference between any two elements from each array.

Solution approach:

1. First, we calculate the sum of absolute differences for each element in nums1 and nums2.

2. We sort nums1 in ascending order.

3. For each element in nums2, we perform a binary search on nums1 to find the closest value to it. We calculate the absolute difference between the two values, subtract it from the previous sum, and store it in a variable.

4. We then update the minimum sum difference if the current sum difference is lower than the previous minimum.

5. Finally, we return the minimum sum difference as the solution.

Code in Python:

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

    # Calculate initial sum of absolute differences
    sum_diff = 0
    for i in range(len(nums1)):
        sum_diff = (sum_diff + abs(nums1[i] - nums2[i])) % mod
    
    # Sort nums1 for binary search
    nums1.sort()
    
    # Perform binary search for closest value in nums1 and calculate new sum of absolute differences
    min_diff = sum_diff
    for i in range(len(nums2)):
        closest_val = bisect.bisect_left(nums1, nums2[i])
        if closest_val > 0:
            new_diff = sum_diff - abs(nums1[closest_val-1] - nums2[i]) + abs(nums1[closest_val] - nums2[i])
            min_diff = min(min_diff, new_diff)
        
        # Check if closest value is the first element in nums1
        if closest_val == 0:
            new_diff = sum_diff - abs(nums1[closest_val] - nums2[i]) + abs(nums1[closest_val+1] - nums2[i])
            min_diff = min(min_diff, new_diff)
    
    return min_diff

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