Range Sum of Sorted Subarray Sums

arraytwo pointersbinary search

Problem Statement

You are given the array nums consisting of n positive integers. You computed the sum of all non-empty continuous subarrays from the array and then sorted them in non-decreasing order, creating a new array of n * (n + 1) / 2 numbers.

Return the sum of the numbers from index left to index right ( indexed from 1 ) , inclusive, in the new array. Since the answer can be a huge number return it modulo 10 ^9 + 7 .

Examples

Example 1

Input: nums = [1,2,3,4], n = 4, left = 1, right = 5
Output: 13
All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13.

Example 2

Input: nums = [1,2,3,4], n = 4, left = 3, right = 4
Output: 6
The given array is the same as example 1. We have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 3 to ri = 4 is 3 + 3 = 6.

Example 3

Input: nums = [1,2,3,4], n = 4, left = 1, right = 10
Output: 50

Constraints

n == nums.length

1 <= nums.length <= 1000

1 <= nums[i] <= 100

1 <= left <= right <= n * (n + 1) / 2

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Medium

Difficulty

Topic Info

ModuleDSA

CategoryArray

Sub-topicTwo Pointers