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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 <= 1051 <= 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:
- The remaining elements only include the prefix of the array $nums$ (which can be empty);
- The remaining elements only include the suffix of the array $nums$;
- 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:
- The remaining elements only include the prefix of the array $nums$ (which can be empty);
- 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:
- $nums[i+1,…,n-1]$;
- $nums[i,…,n-1]$;
- $nums[i-1,…,n-1]$;
- $nums[i-2,…,n-1]$;
- $\cdots$;
- $nums[0,…,n-1]$.
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 $O(1)$.
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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; }