LeetCode in Kotlin
2104. Sum of Subarray Ranges
Medium
You are given an integer array nums. The range of a subarray of nums is the difference between the largest and smallest element in the subarray.
Return the sum of all subarray ranges of nums.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [1,2,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[2], range = 2 - 2 = 0
[3], range = 3 - 3 = 0
[1,2], range = 2 - 1 = 1
[2,3], range = 3 - 2 = 1
[1,2,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 1 + 1 + 2 = 4.
Example 2:
Input: nums = [1,3,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[3], range = 3 - 3 = 0
[3], range = 3 - 3 = 0
[1,3], range = 3 - 1 = 2
[3,3], range = 3 - 3 = 0
[1,3,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 2 + 0 + 2 = 4.
Example 3:
Input: nums = [4,-2,-3,4,1]
Output: 59
Explanation: The sum of all subarray ranges of nums is 59.
Constraints:
1 <= nums.length <= 1000-109 <= nums[i] <= 109
Follow-up: Could you find a solution with O(n) time complexity?
Solution
import java.util.ArrayDeque
import java.util.Deque
class Solution {
fun subArrayRanges(nums: IntArray): Long {
val n = nums.size
var sum: Long = 0
val q: Deque<Int> = ArrayDeque()
q.add(-1)
for (i in 0..n) {
while (q.peekLast() != -1 && (i == n || nums[q.peekLast()] <= nums[i])) {
val cur = q.removeLast()
val left = q.peekLast()
sum += 1L * (cur - left) * (i - cur) * nums[cur]
}
q.add(i)
}
q.clear()
q.add(-1)
for (i in 0..n) {
while (q.peekLast() != -1 && (i == n || nums[q.peekLast()] >= nums[i])) {
val cur = q.removeLast()
val left = q.peekLast()
sum -= 1L * (cur - left) * (i - cur) * nums[cur]
}
q.add(i)
}
return sum
}
}