Today I have done leetcode's August LeetCoding Challenge with cpp.

August LeetCoding Challenge 4

Description

Range Sum of Sorted Subarray Sums

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

class Solution {
  const int MOD = 1e9 + 7;
public:
  int rangeSum(vector<int>& nums, int n, int left, int right) {
    int arr[n * (n + 1) / 2];
    int pos = 0;
    for(int i = 0; i < n; ++i) {
      int s = 0;
      for(int j = i ; j < n; ++j) {
        s += nums[j];
        arr[pos] = s;
        pos += 1;
      }
    }
    sort(arr, arr + n * (n + 1) / 2);
    int answer = 0;
    for(int i = left - 1; i < right; ++i) {
      answer += arr[i];
      answer %= MOD;
    } 
    return answer;
  }
};