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
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.
1 2 3 4 5 6 7 8 9 10 11 |
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 |
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 |
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
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