LeetCode 1685. Sum of Absolute Differences in a Sorted Array

You are given an integer array nums sorted in non-decreasing order.

Build and return an integer array result with the same length as nums such that result[i] is equal to the summation of absolute differences between nums[i] and all the other elements in the array.

In other words, result[i] is equal to sum(|nums[i]-nums[j]|) where 0 <= j < nums.length and j != i (0-indexed).

Example 1:

Input: nums = [2,3,5]

Output: [4,3,5]

Explanation: 

            Assuming the arrays are 0-indexed, then result[0] = |2-2| + |2-3| + |2-5| = 0 + 1 + 3 = 4, result[1] = |3-2| + |3-3| + |3-5| = 1 + 0 + 2 = 3, result[2] = |5-2| + |5-3| + |5-5| = 3 + 2 + 0 = 5.

Example 2:

Input: nums = [1,4,6,8,10]

Output: [24,15,13,15,21]

Constraints:

• 2 <= nums.length <= 105

• 1 <= nums[i] <= nums[i + 1] <= 104

Hide Hint 1

Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted?

Hide Hint 2

For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]).

Hide Hint 3

It can be simplified to (nums[i] * i - (nums[0] + nums[1] + ... + nums[i-1])) + ((nums[i+1] + nums[i+2] + ... + nums[n-1]) - nums[i] * (n-i-1)). One can build prefix and suffix sums to compute this quickly.

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直接按照提示一步一步算就可以了。下面是代码：

Runtime: 4 ms, faster than 62.05% of Java online submissions for Sum of Absolute Differences in a Sorted Array.

Memory Usage: 57.5 MB, less than 77.44% of Java online submissions for Sum of Absolute Differences in a Sorted Array.