#2972

Hard

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Count the Number of Incremovable Subarrays II

Count the number of incremovable subarrays where removal of the subarray results in a strictly increasing array.

Problem Statement

Given a 0-indexed array of positive integers nums, you are asked to find how many subarrays are incremovable. A subarray is called incremovable if, by removing the subarray from nums, the remaining array becomes strictly increasing. For example, in the array [5, 3, 4, 6, 7], the subarray [3, 4] is incremovable because removing it results in [5, 6, 7], which is strictly increasing.

Your goal is to return the total number of such incremovable subarrays within the array nums. You cannot select an empty subarray, and the subarrays must be consecutive. The problem challenges you to find an optimal way of calculating the number of these subarrays efficiently.

Examples

Example 1

Input: nums = [1,2,3,4]

Output: 10

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

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

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

Solution Approach

Binary Search Over Answer Space

The problem can be solved using binary search over the answer space, as you need to find the largest subarray that can be removed while ensuring the remaining array is strictly increasing. This technique leverages a search for the largest valid index where the subsequence from 0 to that index remains strictly increasing.

Two Pointers for Efficient Subarray Check

To efficiently track the valid incremovable subarrays, a two-pointer approach can be employed. The first pointer marks the start of a subarray, while the second pointer moves to expand the valid subarray, ensuring that the subarray removal results in a strictly increasing sequence.

Prefix Array for Increasing Sequences

A prefix array that tracks the longest increasing subsequence up to each index can help in identifying which subarrays, when removed, would maintain the strictly increasing order. This allows for quick determination of valid subarrays.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the binary search and two-pointer approach used to find valid subarrays. In the best case, the time complexity is O(n log n), where n is the length of the input array, due to the binary search and the two-pointer traversal. The space complexity is O(n), as extra space is needed for storing intermediate results such as prefix arrays.

What Interviewers Usually Probe

The candidate understands how to optimize subarray searches using binary search and pointers.
The candidate effectively uses prefix arrays or similar techniques to track increasing sequences.
The candidate can break down a problem requiring multiple steps into a clear, efficient approach.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the problem's constraints, leading to an inefficient solution.
Failing to optimize the subarray search, resulting in excessive time complexity.
Overlooking edge cases where subarrays are close to the entire array, requiring careful handling of boundaries.

Follow-up variants

Increasing the size of the input array to test scalability.
Handling arrays with larger values (e.g., elements in the range of 10^9).
Exploring alternate approaches like dynamic programming or greedy algorithms for subarray validation.

FAQ

What is the primary algorithmic pattern for solving 'Count the Number of Incremovable Subarrays II'?

The problem is typically solved using binary search over the valid answer space, combined with two-pointer techniques to validate subarrays.

How do I know when a subarray is incremovable?

A subarray is incremovable if removing it results in a strictly increasing sequence in the remaining array.

What edge cases should I consider when solving this problem?

Make sure to handle small arrays, large values, and situations where the entire array is considered a valid subarray.

Can I use dynamic programming to solve this problem?

While dynamic programming might be an option, the most efficient approach uses binary search and two pointers to track valid subarrays.

What is the time complexity of the optimal solution?

The optimal solution has a time complexity of O(n log n), where n is the length of the input array, due to the binary search combined with the two-pointer method.

terminal

Solution

Solution 1: Two Pointers

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

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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

Continue Practicing

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