1685. Sum of Absolute Differences in a Sorted Array
Description
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
Solutions
Solution 1: Summation + Enumeration
First, we calculate the sum of all elements in the array $nums$, denoted as $s$. We use a variable $t$ to record the sum of the elements that have been enumerated so far.
Next, we enumerate $nums[i]$. Then $ans[i] = nums[i] \times i - t + s - t - nums[i] \times (n - i)$. After that, we update $t$, i.e., $t = t + nums[i]$. We continue to enumerate the next element until all elements are enumerated.
The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
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