Medium
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
class Solution {
fun rangeSum(nums: IntArray, n: Int, left: Int, right: Int): Int {
val len = n * (n + 1) / 2
val arr = IntArray(len)
var idx = 0
var prev = 0
for (i in 0 until n) {
for (j in i until n) {
arr[idx] = prev + nums[j]
prev = arr[idx]
idx++
}
prev = 0
}
arr.sort()
var result = 0
val mod = 1000000007
for (i in left - 1 until right) {
result = (result + arr[i]) % mod
}
return result
}
}