1508. Range Sum of Sorted Subarray Sums

Description

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

range-sum-of-sorted-subarray-sums.py

class Solution:
    def rangeSum(self, nums: List[int], n: int, left: int, right: int) -> int:
        n = len(nums)
        arr = []
        M = int(1e9+7)

        for i in range(n):
            s = nums[i]
            for j in range(i+1,n):
                s += nums[j]
                nums.append(s)

        nums = sorted(nums)
        res = 0
        for i in range(left-1, right):
            res = (res + nums[i])%M

        return res%M

range-sum-of-sorted-subarray-sums.cpp

class Solution {
public:
    int rangeSum(vector<int>& nums, int n, int left, int right) {
        vector<int> v;

        for (int i = 0; i < n; ++i){
            int s = 0;
            for (int j = i; j < n; ++j){
                s += nums[j];
                v.push_back(s);

            }
        }

        sort(v.begin(),v.end());

        --left, --right;

        int ans = 0, M = 1e9+7;
        for (int i = left; i <= right; ++i){
            ans = (ans+v[i])%M;
        }

        return ans;

    }
};