1508.Range Sum of Sorted Subarray Sums

Tags: Medium Array Sort

Links: https://leetcode.com/problems/range-sum-of-sorted-subarray-sums/

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Given the array nums consisting of n positive integers. You computed the sum of all non-empty continous subarrays from the array and then sort 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:

• 1 <= nums.length <= 10^3
• nums.length == n
• 1 <= nums[i] <= 100
• 1 <= left <= right <= n * (n + 1) / 2

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计算连续区间和第一时间想到前缀和，然后就是计算n * (n + 1) / 2个子数组的区间和的值，存入res数组。将res排序，计算下标从left - 1到right - 1的和即可，注意取模。

时间复杂度$O(n^2\log{n})$

class Solution {
public:
    int rangeSum(vector<int>& nums, int n, int left, int right) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        vector<int> preSum(n + 1, 0);
        for (int i = 1; i <= n; ++i) preSum[i] = preSum[i - 1] + nums[i - 1];

        vector<int> res;
        for (int i = 1; i <= n; ++i) {
            for (int j = i; j <= n; ++j) {
                res.push_back(preSum[j] - preSum[i - 1]);
            }
        }

        sort(res.begin(), res.end());
        const int MODE = 1e9 + 7;

        int ans = 0;
        for (int i = left - 1; i < right; ++i) ans = (ans + res[i]) % MODE;

        return ans;
    }
};