human
I found an interesting problem on https://leetcode.com/problems/range-sum-of-sorted-subarray-sums/:
Given the array nums consisting of n positive integers. You computed the sum of all non-empty continous subarrays from the array and then sort 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.
 
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:

1 <= nums.length <= 10^3
nums.length == n
1 <= nums[i] <= 100
1 <= left <= right <= n * (n + 1) / 2

I tried it in Python, but could not do it. Can you solve it?

Can you provide a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
    def rangeSum(self, nums: List[int], n: int, left: int, right: int) -> int:
        s = [0]
        for x in nums:
            s.append(s[-1] + x)
        n = len(nums)
        q = [(x, i, i) for i, x in enumerate(nums)]
        heapq.heapify(q)
        i = 1
        ans = 0
        while i <= right:
            x, j, k = heapq.heappop(q)
            if i >= left:
                ans += x
            if k > 0:
                k -= 1
                y = s[j + 1] - s[k]
                heapq.heappush(q,(y, j, k))
            i += 1
        return ans % (10 ** 9 + 7)