Package g1501_1600.s1508_range_sum_of_sorted_subarray_sums

Class Solution

java.lang.Object

g1501_1600.s1508_range_sum_of_sorted_subarray_sums.Solution

public class Solution
extends Object

1508 - Range Sum of Sorted Subarray Sums\. Medium 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`

Constructor Summary

Constructors

Constructor	Description
Solution()

Method Summary

Modifier and Type	Method	Description
int	rangeSum(int[] nums, int n, int left, int right)

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Details

Solution

public Solution()

Method Details

rangeSum

public int rangeSum(int[] nums,
 int n,
 int left,
 int right)