How do you solve Minimum Absolute Sum Difference on LeetCode?

Minimum Absolute Sum Difference (LeetCode #1818) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Replacing an element in nums1 can significantly reduce the absolute sum difference.

What is the brute force solution for Minimum Absolute Sum Difference?

The brute force approach for Minimum Absolute Sum Difference is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible replacement of an element in nums1 with every other element in nums1. This ensures we find the minimum absolute sum difference, but it's inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Absolute Sum Difference?

Minimum Absolute Sum Difference uses the following concepts: Array, Binary Search, Sorting, Ordered Set. The recommended patterns to study are: Binary Search, Sorting.

What are the interview tips for Minimum Absolute Sum Difference?

Always start with a brute force solution to understand the problem better. Think about how you can reduce the problem size or complexity through sorting or hashing. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Absolute Sum Difference?

Not considering edge cases where nums1 and nums2 are already equal. Failing to optimize the search for the closest element in nums1.

#1818

Minimum Absolute Sum Difference

Medium

ArrayBinary SearchSortingOrdered SetBinary SearchSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal approach leverages sorting and binary search to efficiently find the best replacement for each element in nums1. This reduces the time complexity significantly while ensuring we still minimize the absolute sum difference.

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Algorithm

4 steps

1Step 1: Calculate the initial absolute sum difference between nums1 and nums2.
2Step 2: Sort nums1 to enable binary search.
3Step 3: For each element in nums2, use binary search to find the closest value in nums1 that minimizes the absolute difference.
4Step 4: Calculate the new absolute sum difference for each replacement and keep track of the minimum.

solution.py16 lines

1def minAbsoluteSumDiff(nums1, nums2):
2    import bisect
3    mod = 10**9 + 7
4    n = len(nums1)
5    initial_sum = sum(abs(nums1[i] - nums2[i]) for i in range(n))
6    nums1_sorted = sorted(nums1)
7    min_sum = initial_sum
8    for num in nums2:
9        idx = bisect.bisect_left(nums1_sorted, num)
10        if idx < len(nums1_sorted):
11            new_sum = initial_sum - abs(num - nums1[idx]) + abs(num - nums1_sorted[idx])
12            min_sum = min(min_sum, new_sum)
13        if idx > 0:
14            new_sum = initial_sum - abs(num - nums1[idx-1]) + abs(num - nums1_sorted[idx-1])
15            min_sum = min(min_sum, new_sum)
16    return min_sum % mod

ℹ

Complexity note: The time complexity is O(n log n) due to sorting nums1, and O(n) for iterating through nums2 and performing binary searches. The space complexity is O(n) for storing the sorted version of nums1.

1Replacing an element in nums1 can significantly reduce the absolute sum difference.
2Sorting nums1 allows for efficient searching of the closest elements to minimize differences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.