1508. Range Sum of Sorted Subarray Sums

/*
 * @lc app=leetcode id=1508 lang=cpp
 *
 * [1508] Range Sum of Sorted Subarray Sums
 *
 * https://leetcode.com/problems/range-sum-of-sorted-subarray-sums/description/
 *
 * algorithms
 * Medium (63.14%)
 * Likes:    1557
 * Dislikes: 262
 * Total Accepted:    179.2K
 * Total Submissions: 284K
 * Testcase Example:  '[1,2,3,4]\n4\n1\n5'
 *
 * 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
 *
 *
 */

// @lc code=start
class Solution {
public:
    int rangeSum(vector<int>& nums, int n, int left, int right) {
        vector<int> rangeSum;
        for (int i = 0; i < nums.size(); i++) {
            int sum = 0;
            for (int j = i; j < nums.size(); j++) {
                sum += nums[j];
                rangeSum.push_back(sum);
            }
        }
        sort(rangeSum.begin(), rangeSum.end());
        int res = 0;
        const int MOD = 1e9 + 7;
        for (int i = left - 1; i < right; i++) {
            res = (res + rangeSum[i]) % MOD;
        }
        return res;
    }
};
// @lc code=end

• T: $O(n \log sum)$
• S: $O(1)$