1685. Sum of Absolute Differences in a Sorted Array

Medium

Solution

The initial equation is

for (i in nums.indices) {
    result[i] = (i until n).sumOf { nums[it] } - (0 until i).sumOf { nums[it] } + nums[i] * (i * 2 - n)
}

Can be simplified to

var sumByCurrent = nums.sum()
var sumFuture = 0

for (i in nums.indices) {
    result[i] = sumByCurrent - sumFuture + nums[i] * (i * 2 - n)
    sumByCurrent -= nums[i]
    sumFuture += nums[i]
}

And finally,

var sumByCurrent = nums.sum()

for (i in nums.indices) {
    result[i] = sumByCurrent + nums[i] * (i * 2 - n)
    sumByCurrent -= nums[i] * 2
}

O(N) time and O(N) space.

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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