Range_Sum_of_Sorted_Subarray_Sums.cpp

// Range Sum of Sorted Subarray Sums leetcode 1508

// 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

int rangeSum(vector<int>& nums, int n, int left, int right) 
{
    vector<int> temp;

    for(int i = 0 ; i < n ; i++)
    {   
        int sum = 0 ;
        for(int j=i;j<n;j++)
        {
            sum+=nums[j];
            temp.push_back(sum);
        }
    }

    sort(temp.begin(),temp.end());
    long ans=0;
    
    for(int i = left-1 ; i < right;i++)
    {
        ans+=temp[i];
    }   

    return ans%1000000007;
}