# 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

Solution:

```class Solution {
static class Pair {
long sum;
int index;

Pair(long sum, int index) {
this.sum = sum;
this.index = index;
}
}

public int rangeSum(int[] nums, int n, int left, int right) {
PriorityQueue<Pair> minHeap  = new PriorityQueue<Pair> (n, (p1, p2) -> Long.compare(p1.sum, p2.sum));
for (int i = 0; i < nums.length; i ++) {
minHeap.offer(new Pair(nums[i], i + 1));
}
int sum = 0, mod = (int) 1e9 + 7;
for (int i = 1; i <= right; i ++) {
Pair curr = minHeap.poll();
if (i >= left) {
sum = (int) ((sum + curr.sum) % mod);
}
if (curr.index < n) {
curr.sum += nums[curr.index++];
minHeap.offer(curr);
}
}
return sum;
}
}```