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:


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;
    }
}