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2104. Sum of Subarray Ranges

Difficulty Topics

Description

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?

Solution

sum-of-subarray-ranges.py
from sortedcontainers import SortedList

class Solution:
    def subArrayRanges(self, nums: List[int]) -> int:
        res = 0
        n = len(nums)

        for i in range(n):
            smallest = largest = nums[i]
            for j in range(i + 1, n):
                smallest = min(smallest, nums[j])
                largest = max(largest, nums[j])

                res += largest - smallest

        return res