• Home
• com.github.javadev
• leetcode-in-java21
• 1.30
• source code
• readme.md

×

Project download

Many resources are needed to download a project. Please understand that we have to compensate our server costs. Thank you in advance.
Project price only 1 $

You can buy this project and download/modify it how often you want.

g2101_2200.s2104_sum_of_subarray_ranges.readme.md Maven / Gradle / Ivy

2104\. Sum of Subarray Ranges

Medium

You are given an integer array `nums`. The **range** of a subarray of `nums` is the difference between the largest and smallest element in the subarray.

Return _the **sum of all** subarray ranges of_ `nums`_._

A subarray is a contiguous **non-empty** sequence of elements within an array.

**Example 1:**

**Input:** nums = [1,2,3]

**Output:** 4

**Explanation:** The 6 subarrays of nums are the following: 

[1], range = largest - smallest = 1 - 1 = 0 

[2], range = 2 - 2 = 0 

[3], range = 3 - 3 = 0 

[1,2], range = 2 - 1 = 1 

[2,3], range = 3 - 2 = 1 

[1,2,3], range = 3 - 1 = 2 

So the sum of all ranges is 0 + 0 + 0 + 1 + 1 + 2 = 4.

**Example 2:**

**Input:** nums = [1,3,3]

**Output:** 4

**Explanation:** The 6 subarrays of nums are the following: 

[1], range = largest - smallest = 1 - 1 = 0 

[3], range = 3 - 3 = 0 

[3], range = 3 - 3 = 0 

[1,3], range = 3 - 1 = 2 

[3,3], range = 3 - 3 = 0 

[1,3,3], range = 3 - 1 = 2 

So the sum of all ranges is 0 + 0 + 0 + 2 + 0 + 2 = 4.

**Example 3:**

**Input:** nums = [4,-2,-3,4,1]

**Output:** 59

**Explanation:** The sum of all subarray ranges of nums is 59.

**Constraints:**

*   `1 <= nums.length <= 1000`
*   -109 <= nums[i] <= 109

**Follow-up:** Could you find a solution with `O(n)` time complexity?