1615 Range Sum Of Sorted Subarray Sums

Range Sum of Sorted Subarray Sums

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 109 + 7.

 

Example 1:

**Input:** nums = [1,2,3,4], n = 4, left = 1, right = 5
**Output:** 13 
**Explanation:** 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
**Explanation:** 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

class Solution:
    def rangeSum(self, nums: List[int], n: int, left: int, right: int) -> int:
        i = 0
        j = 0
        arr = []
        while i < len(nums):
            j = i
            sum1 = 0
            while j < len(nums):
                sum1 += nums[j]
                arr.append(sum1)
                j += 1
            i += 1
        
        arr.sort()
        k = arr[left-1:right]
        x = sum(k)
        return x %  (pow(10,9) + 7)