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.
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.
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.
Input: nums = [1,2,3,4], n = 4, left = 1, right = 10 Output: 50
1 <= nums.length <= 10^3
nums.length == n
1 <= nums[i] <= 100
1 <= left <= right <= n * (n + 1) / 2
# @param {Integer[]} nums
# @param {Integer} n
# @param {Integer} left
# @param {Integer} right
# @return {Integer}
def range_sum(nums, _n, left, right)
sums = []
ret = 0
(0...nums.size).each do |i|
sum = 0
(i...nums.size).each do |j|
sum += nums[j]
sums.push(sum)
end
end
sums.sort!
(left..right).each do |i|
ret = (ret + sums[i - 1]) % 1_000_000_007
end
ret
end
impl Solution {
pub fn range_sum(nums: Vec<i32>, n: i32, left: i32, right: i32) -> i32 {
let mut sums = vec![];
let mut ret = 0;
for i in 0..nums.len() {
let mut sum = 0;
for j in i..nums.len() {
sum += nums[j];
sums.push(sum);
}
}
sums.sort_unstable();
for i in left..=right {
ret = (ret + sums[i as usize - 1]) % 1_000_000_007;
}
ret
}
}