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 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:

• n == nums.length

• 1 <= nums.length <= 1000

• 1 <= nums[i] <= 100

• 1 <= left <= right <= n * (n + 1) / 2

Solution (Java)

class Solution {
    public int rangeSum(int[] nums, int n, int left, int right) {
        int len = n * (n + 1) / 2;
        int[] arr = new int[len];
        int idx = 0;
        int prev = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                arr[idx] = prev + nums[j];
                prev = arr[idx];
                idx++;
            }
            prev = 0;
        }
        Arrays.sort(arr);
        int result = 0;
        int mod = 1000000007;
        for (int i = left - 1; i < right; i++) {
            result = (result + arr[i]) % mod;
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).