README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1600-1699/1685.Sum%20of%20Absolute%20Differences%20in%20a%20Sorted%20Array/README_EN.md	1495	Biweekly Contest 41 Q2	Array Math Prefix Sum

1685. Sum of Absolute Differences in a Sorted Array

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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 $O(1)$.

Python3

class Solution:
    def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:
        ans = []
        s, t = sum(nums), 0
        for i, x in enumerate(nums):
            v = x * i - t + s - t - x * (len(nums) - i)
            ans.append(v)
            t += x
        return ans

Java

class Solution {
    public int[] getSumAbsoluteDifferences(int[] nums) {
        // int s = Arrays.stream(nums).sum();
        int s = 0, t = 0;
        for (int x : nums) {
            s += x;
        }
        int n = nums.length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int v = nums[i] * i - t + s - t - nums[i] * (n - i);
            ans[i] = v;
            t += nums[i];
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> getSumAbsoluteDifferences(vector<int>& nums) {
        int s = accumulate(nums.begin(), nums.end(), 0), t = 0;
        int n = nums.size();
        vector<int> ans(n);
        for (int i = 0; i < n; ++i) {
            int v = nums[i] * i - t + s - t - nums[i] * (n - i);
            ans[i] = v;
            t += nums[i];
        }
        return ans;
    }
};

Go

func getSumAbsoluteDifferences(nums []int) (ans []int) {
	var s, t int
	for _, x := range nums {
		s += x
	}
	for i, x := range nums {
		v := x*i - t + s - t - x*(len(nums)-i)
		ans = append(ans, v)
		t += x
	}
	return
}

TypeScript

function getSumAbsoluteDifferences(nums: number[]): number[] {
    const s = nums.reduce((a, b) => a + b);
    let t = 0;
    const n = nums.length;
    const ans = new Array(n);
    for (let i = 0; i < n; ++i) {
        const v = nums[i] * i - t + s - t - nums[i] * (n - i);
        ans[i] = v;
        t += nums[i];
    }
    return ans;
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var getSumAbsoluteDifferences = function (nums) {
    const s = nums.reduce((a, b) => a + b);
    let t = 0;
    const n = nums.length;
    const ans = new Array(n);
    for (let i = 0; i < n; ++i) {
        const v = nums[i] * i - t + s - t - nums[i] * (n - i);
        ans[i] = v;
        t += nums[i];
    }
    return ans;
};

C#

public class Solution {
    public int[] GetSumAbsoluteDifferences(int[] nums) {
        int s = 0, t = 0;
        foreach (int x in nums) {
            s += x;
        }
        int n = nums.Length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int v = nums[i] * i - t + s - t - nums[i] * (n - i);
            ans[i] = v;
            t += nums[i];
        }
        return ans;
    }
}