#2970

Easy

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Count the Number of Incremovable Subarrays I

Count the number of incremovable subarrays in an array by removing them to create a strictly increasing sequence.

array two pointers binary search enumeration

Problem Statement

You are given a 0-indexed array of positive integers nums. A subarray is called incremovable if removing it results in a strictly increasing sequence. For example, in the array [5, 3, 4, 6, 7], the subarray [3, 4] is incremovable because removing it gives the array [5, 6, 7], which is strictly increasing.

Return the total number of incremovable subarrays in nums. A subarray is defined as a contiguous segment of the array, and you must remove at least one element. An empty subarray is not allowed.

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 <= 50
1 <= nums[i] <= 50

Solution Approach

Binary Search for Answer Space

We can use binary search over the valid subarray removal space. By narrowing down the range for possible removals, we can efficiently determine which subarrays will make the array strictly increasing.

Two Loops for Subarray Checking

To validate each subarray, use two nested loops to check all potential subarrays. For each subarray, we simulate removing it and check if the remaining array is strictly increasing.

Optimize with Early Termination

As we check each subarray, we can stop early if the remaining array fails to become strictly increasing. This helps to reduce unnecessary computations and improves overall performance.

Complexity Analysis

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

The time complexity depends on the number of subarrays checked, which can be up to O(n^2). The binary search optimization can reduce the search space, but the primary time cost comes from iterating through subarrays. Space complexity is O(n) due to the need for temporary storage of subarrays during validation.

What Interviewers Usually Probe

Does the candidate use binary search effectively to optimize the subarray check process?
Can the candidate explain the early termination technique when validating subarrays?
Is the candidate aware of how to optimize the solution for larger arrays by reducing unnecessary checks?

Common Pitfalls or Variants

Common pitfalls

Failing to recognize the need for an efficient subarray check can lead to an O(n^3) solution.
Not using binary search correctly, which could make the solution inefficient for larger inputs.
Ignoring the early termination condition when checking subarrays, resulting in excessive computation time.

Follow-up variants

Extend the problem to handle arrays with non-positive integers.
Consider a version where removing multiple non-contiguous subarrays can also make the array strictly increasing.
Test the solution for larger input sizes, such as arrays with 100 elements, to measure performance.

FAQ

What is the main approach to solve 'Count the Number of Incremovable Subarrays I'?

The problem can be solved efficiently by using binary search over the answer space combined with two loops to check all possible subarrays.

How does binary search help in this problem?

Binary search helps optimize the process by narrowing down the valid subarray removal range, reducing unnecessary checks and improving efficiency.

What does it mean for a subarray to be incremovable?

An incremovable subarray is one that, when removed, leaves the remaining array strictly increasing without any violations of order.

What is the time complexity of solving this problem?

The time complexity is typically O(n^2) due to the need to check all subarrays, but optimizations like binary search can help reduce unnecessary checks.

What are common mistakes when solving this problem?

Common mistakes include failing to use binary search effectively, not terminating early when the subarray validation fails, and checking subarrays inefficiently with O(n^3) time complexity.

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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