1508. Range Sum of Sorted Subarray Sums

Problem

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

Solution

/**
 * @param {number[]} nums
 * @param {number} n
 * @param {number} left
 * @param {number} right
 * @return {number}
 */
var rangeSum = function(nums, n, left, right) {
    var queue = new MinPriorityQueue();
    for (var i = 0; i < nums.length; i++) {
        queue.enqueue(i, nums[i]);
    }
    var res = 0;
    var mod = Math.pow(10, 9) + 7;
    for (var j = 0; j < right; j++) {
        var { element, priority } = queue.dequeue();
        if (element < nums.length - 1) {
            queue.enqueue(element + 1, priority + nums[element + 1]);
        }
        if (j >= left - 1) {
            res += priority;
            res %= mod;
        }
    }
    return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n^2 * log(n)).
• Space complexity : O(n).