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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

Solutions

Solution 1: Simulation

Thinking

We must sort every subarray sum and add the entries from index \(\textit{left}\) to \(\textit{right}\). There are \(O(n^2)\) subarrays and \(n\le 10^3\). Recomputing that list per query would waste work, but here there is a single query.

Enumerate the left endpoint and accumulate a running sum to the right to list all subarray sums in \(O(n^2)\). After sorting, sum the requested closed range and reduce modulo \(10^9+7\). An \(n^2\log n\) sort fits the limits.

We can generate the array \(\textit{arr}\) according to the problem's requirements, then sort the array, and finally calculate the sum of all elements in the range \([\textit{left}-1, \textit{right}-1]\) to get the result.

The time complexity is \(O(n^2 \times \log n)\), and the space complexity is \(O(n^2)\). Here, \(n\) is the length of the array given in the problem.

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class Solution:
    def rangeSum(self, nums: List[int], n: int, left: int, right: int) -> int:
        arr = []
        for i in range(n):
            s = 0
            for j in range(i, n):
                s += nums[j]
                arr.append(s)
        arr.sort()
        mod = 10**9 + 7
        return sum(arr[left - 1 : right]) % mod

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class Solution {
    public int rangeSum(int[] nums, int n, int left, int right) {
        int[] arr = new int[n * (n + 1) / 2];
        for (int i = 0, k = 0; i < n; ++i) {
            int s = 0;
            for (int j = i; j < n; ++j) {
                s += nums[j];
                arr[k++] = s;
            }
        }
        Arrays.sort(arr);
        int ans = 0;
        final int mod = (int) 1e9 + 7;
        for (int i = left - 1; i < right; ++i) {
            ans = (ans + arr[i]) % mod;
        }
        return ans;
    }
}

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

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func rangeSum(nums []int, n int, left int, right int) (ans int) {
    var arr []int
    for i := 0; i < n; i++ {
        s := 0
        for j := i; j < n; j++ {
            s += nums[j]
            arr = append(arr, s)
        }
    }
    sort.Ints(arr)
    const mod int = 1e9 + 7
    for _, x := range arr[left-1 : right] {
        ans = (ans + x) % mod
    }
    return
}

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function rangeSum(nums: number[], n: number, left: number, right: number): number {
    let arr = Array((n * (n + 1)) / 2).fill(0);
    const mod = 10 ** 9 + 7;

    for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
        for (let j = i; j < n; j++, k++) {
            s += nums[j];
            arr[k] = s;
        }
    }

    arr = arr.sort((a, b) => a - b).slice(left - 1, right);
    return arr.reduce((acc, cur) => (acc + cur) % mod, 0);
}

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function rangeSum(nums, n, left, right) {
    let arr = Array((n * (n + 1)) / 2).fill(0);
    const mod = 10 ** 9 + 7;

    for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
        for (let j = i; j < n; j++, k++) {
            s += nums[j];
            arr[k] = s;
        }
    }

    arr = arr.sort((a, b) => a - b).slice(left - 1, right);
    return arr.reduce((acc, cur) => acc + cur, 0) % mod;
}

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