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.length1 <= nums.length <= 10001 <= nums[i] <= 1001 <= left <= right <= n * (n + 1) / 2
Solutions
Solution 1: Simulation
We can generate the array arr according to the problem's requirements, then sort the array, and finally calculate the sum of all elements in the range [left-1, right-1] to get the result.
The time complexity is O(n2 × log n), and the space complexity is O(n2). Here, n is the length of the array given in the problem.
PythonJavaC++GoTypeScriptJavaScript
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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);
}(code-box)
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;
}(code-box)
