LeetCode in Kotlin
2972. Count the Number of Incremovable Subarrays II
Hard
You are given a 0-indexed array of positive integers nums.
A subarray of nums is called incremovable if nums becomes strictly increasing on removing the subarray. For example, the subarray [3, 4] is an incremovable subarray of [5, 3, 4, 6, 7] because removing this subarray changes the array [5, 3, 4, 6, 7] to [5, 6, 7] which is strictly increasing.
Return the total number of incremovable subarrays of nums.
Note that an empty array is considered strictly increasing.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [1,2,3,4]
Output: 10
Explanation: The 10 incremovable subarrays are: [1], [2], [3], [4], [1,2], [2,3], [3,4], [1,2,3], [2,3,4], and [1,2,3,4], because on removing any one of these subarrays nums becomes strictly increasing. Note that you cannot select an empty subarray.
Example 2:
Input: nums = [6,5,7,8]
Output: 7
Explanation: The 7 incremovable subarrays are: [5], [6], [5,7], [6,5], [5,7,8], [6,5,7] and [6,5,7,8]. It can be shown that there are only 7 incremovable subarrays in nums.
Example 3:
Input: nums = [8,7,6,6]
Output: 3
Explanation: The 3 incremovable subarrays are: [8,7,6], [7,6,6], and [8,7,6,6]. Note that [8,7] is not an incremovable subarray because after removing [8,7] nums becomes [6,6], which is sorted in ascending order but not strictly increasing.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 109
Solution
class Solution {
fun incremovableSubarrayCount(nums: IntArray): Long {
var ans: Long
val n = nums.size
var l = 0
var r = n - 1
while (l + 1 < n && nums[l] < nums[l + 1]) {
l++
}
while (r > 0 && nums[r - 1] < nums[r]) {
r--
}
ans = (if ((l == n - 1)) 0 else 1 + (n - r)).toLong()
for (i in 0..l) {
while (r < n && nums[r] <= nums[i]) {
r++
}
ans += (n - r + 1).toLong()
}
return ans
}
}