#1818

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Minimum Absolute Sum Difference

Minimize the absolute sum difference between two integer arrays by replacing at most one element from the first array.

Problem Statement

You are given two arrays of positive integers, nums1 and nums2, both of length n. The absolute sum difference between the arrays is defined as the sum of |nums1[i] - nums2[i]| for all 0 <= i < n. You can replace at most one element of nums1 with any other element in nums1 to minimize this absolute sum difference.

Your task is to find the minimum possible absolute sum difference after performing at most one replacement in nums1. The solution requires considering the best potential replacement for each element in nums1 using binary search and sorting strategies.

Examples

Example 1

Input: nums1 = [1,7,5], nums2 = [2,3,5]

Output: 3

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

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

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

Solution Approach

Binary Search over Answer Space

The key to solving this problem efficiently lies in using binary search over the potential answer space. For each element in nums1, determine the optimal replacement that minimizes the absolute sum difference with the corresponding element in nums2.

Sorting and Searching

Sort nums1 and nums2, then use binary search to find the element in nums1 that, when replaced, minimizes the sum of absolute differences. This approach leverages sorting to speed up the search for potential replacements.

Greedy Testing of Replacements

After sorting, test each possible replacement for the elements in nums1, compute the resulting absolute sum difference, and keep track of the minimum value found. This greedy approach ensures that the best possible replacement is chosen.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity primarily depends on the sorting steps and binary search operations. Sorting nums1 takes O(n log n), and performing binary search for each element adds another O(log n) per element, leading to a total time complexity of O(n log n). Space complexity is O(n) due to the storage of the arrays and auxiliary data structures.

What Interviewers Usually Probe

Look for an understanding of binary search over a sorted array.
Test candidate's ability to identify and implement the optimal replacement strategy efficiently.
Gauge how well the candidate handles edge cases and the upper constraint limits.

Common Pitfalls or Variants

Common pitfalls

Failing to account for the case where no replacement is needed (when nums1 is already equal to nums2).
Incorrectly implementing binary search, leading to suboptimal performance or incorrect answers.
Overcomplicating the problem by using unnecessary operations instead of relying on sorting and binary search.

Follow-up variants

Generalize the problem by allowing multiple replacements in nums1 to minimize the absolute sum difference.
Modify the problem to handle larger constraints, such as arrays with length up to 10^6.
Introduce the requirement of minimizing the absolute sum difference for non-positive integers in nums1 and nums2.

FAQ

What is the main approach to solve Minimum Absolute Sum Difference?

The main approach is to use binary search over the valid answer space, sorting the arrays and testing the best replacement for each element in nums1.

What is the time complexity of the solution?

The time complexity is O(n log n) due to the sorting of nums1 and the binary search operations for each element.

Can the problem be solved without sorting?

Sorting is necessary to efficiently perform binary search and find the best replacement for each element in nums1.

What is the role of binary search in this problem?

Binary search helps to find the best element from nums1 that minimizes the absolute sum difference when tested against each element of nums2.

What are some common mistakes when solving this problem?

Common mistakes include failing to handle cases where no replacement is needed and incorrect binary search implementation leading to inefficient or incorrect solutions.

terminal

Solution

Solution 1: Sorting + Binary Search

According to the problem, we can first calculate the absolute difference sum of `nums1` and `nums2` without any replacements, denoted as $s$.

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class Solution:
    def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
        mod = 10**9 + 7
        nums = sorted(nums1)
        s = sum(abs(a - b) for a, b in zip(nums1, nums2)) % mod
        mx = 0
        for a, b in zip(nums1, nums2):
            d1, d2 = abs(a - b), inf
            i = bisect_left(nums, b)
            if i < len(nums):
                d2 = min(d2, abs(nums[i] - b))
            if i:
                d2 = min(d2, abs(nums[i - 1] - b))
            mx = max(mx, d1 - d2)
        return (s - mx + mod) % mod

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