2972. Count the Number of Incremovable Subarrays II

Description

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 <= 105
• 1 <= nums[i] <= 109

Solutions

Solution 1: Two Pointers

According to the problem description, after removing a subarray, the remaining elements are strictly increasing. Therefore, there are several situations:

1. The remaining elements only include the prefix of the array $nums$ (which can be empty);
2. The remaining elements only include the suffix of the array $nums$;
3. The remaining elements include both the prefix and the suffix of the array $nums$.

The second and third situations can be combined into one, that is, the remaining elements include the suffix of the array $nums$. So there are two situations in total:

1. The remaining elements only include the prefix of the array $nums$ (which can be empty);
2. The remaining elements include the suffix of the array $nums$.

First, consider the first situation, that is, the remaining elements only include the prefix of the array $nums$. We can use a pointer $i$ to point to the last element of the longest increasing prefix of the array $nums$, that is, $nums[0] \lt nums[1] \lt \cdots \lt nums[i]$, then the number of remaining elements is $n - i - 1$, where $n$ is the length of the array $nums$. Therefore, for this situation, to make the remaining elements strictly increasing, we can choose to remove the following subarrays:

1. $nums[i+1,…,n-1]$;
2. $nums[i,…,n-1]$;
3. $nums[i-1,…,n-1]$;
4. $nums[i-2,…,n-1]$;
5. $\cdots$;
6. $nums[0,…,n-1]$.

There are $i + 2$ situations in total, so for this situation, the number of removed increasing subarrays is $i + 2$.

Next, consider the second situation, that is, the remaining elements include the suffix of the array $nums$. We can use a pointer $j$ to point to the first element of the increasing suffix of the array $nums$. We enumerate $j$ as the first element of the increasing suffix in the range $[n - 1,…,1]$. Each time, we need to move the pointer $i$ to make $nums[i] \lt nums[j]$, then the number of removed increasing subarrays increases by $i + 2$. When $nums[j - 1] \ge nums[j]$, we stop enumerating because the suffix is not strictly increasing at this time.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

• class Solution {
public long incremovableSubarrayCount(int[] nums) {
int i = 0, n = nums.length;
while (i + 1 < n && nums[i] < nums[i + 1]) {
++i;
}
if (i == n - 1) {
return n * (n + 1L) / 2;
}
long ans = i + 2;
for (int j = n - 1; j > 0; --j) {
while (i >= 0 && nums[i] >= nums[j]) {
--i;
}
ans += i + 2;
if (nums[j - 1] >= nums[j]) {
break;
}
}
return ans;
}
}

• class Solution {
public:
long long incremovableSubarrayCount(vector<int>& nums) {
int i = 0, n = nums.size();
while (i + 1 < n && nums[i] < nums[i + 1]) {
++i;
}
if (i == n - 1) {
return n * (n + 1LL) / 2;
}
long long ans = i + 2;
for (int j = n - 1; j > 0; --j) {
while (i >= 0 && nums[i] >= nums[j]) {
--i;
}
ans += i + 2;
if (nums[j - 1] >= nums[j]) {
break;
}
}
return ans;
}
};

• class Solution:
def incremovableSubarrayCount(self, nums: List[int]) -> int:
i, n = 0, len(nums)
while i + 1 < n and nums[i] < nums[i + 1]:
i += 1
if i == n - 1:
return n * (n + 1) // 2
ans = i + 2
j = n - 1
while j:
while i >= 0 and nums[i] >= nums[j]:
i -= 1
ans += i + 2
if nums[j - 1] >= nums[j]:
break
j -= 1
return ans


• func incremovableSubarrayCount(nums []int) int64 {
i, n := 0, len(nums)
for i+1 < n && nums[i] < nums[i+1] {
i++
}
if i == n-1 {
return int64(n * (n + 1) / 2)
}
ans := int64(i + 2)
for j := n - 1; j > 0; j-- {
for i >= 0 && nums[i] >= nums[j] {
i--
}
ans += int64(i + 2)
if nums[j-1] >= nums[j] {
break
}
}
return ans
}

• function incremovableSubarrayCount(nums: number[]): number {
const n = nums.length;
let i = 0;
while (i + 1 < n && nums[i] < nums[i + 1]) {
i++;
}
if (i === n - 1) {
return (n * (n + 1)) / 2;
}
let ans = i + 2;
for (let j = n - 1; j; --j) {
while (i >= 0 && nums[i] >= nums[j]) {
--i;
}
ans += i + 2;
if (nums[j - 1] >= nums[j]) {
break;
}
}
return ans;
}