[Leetcode] Problem 1508 - Range Sum of Sorted Subarray Sums

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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

No.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.

No.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.

No.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

Code

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public int rangeSum(int[] nums, int n, int left, int right) {
    int result = 0;
    int mod = (int) 1e9 + 7;
    PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);

    for (int i = 0; i < nums.length; i++)
        pq.offer(new int[] {nums[i] , i});

    for (int i = 1; i <= right; i++) {
        int[] array = pq.poll();
        int sum = array[0];
        int idx = array[1];

        if (i >= left)
            result = (result + sum) % mod;

        if (idx < n - 1)
            pq.offer(new int[] {sum + nums[idx + 1], idx + 1});
    }

    return result;
}