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

## Solution: Prefix Sum

Let s[i] denote sum(num[i] – num[j]) 0 <= j <= i
s[i] = s[i – 1] + (num[i] – num[i – 1]) * i
Let l[i] denote sum(nums[j] – nums[i]) i <= j < n
l[i] = l[i + 1] + (nums[i + 1] – num[i]) * (n – i – 1)
ans[i] = s[i] + l[i]

e.g. 1, 3, 7, 9
s[0] = 0
s[1] = 0 + (3 – 1) * 1 = 2
s[2] = 2 + (7 – 3) * 2 = 10
s[3] = 10 + (9 – 7) * 3 = 16
l[3] = 0
l[2] = 0 + (9 – 7) * 1 = 2
l[1] = 2 + (7 – 3) * 2 = 10
l[0] = 10 + (3 – 1) * 3 = 16

ans = [16, 12, 12, 16]

Time complexity: O(n)
Space complexity: O(n)

## C++

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