How do you solve Count the Number of Incremovable Subarrays II on LeetCode?

Count the Number of Incremovable Subarrays II (LeetCode #2972) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Identifying boundaries of strictly increasing sequences is crucial.

What is the time complexity of Count the Number of Incremovable Subarrays II?

The optimal solution for Count the Number of Incremovable Subarrays II runs in O(n) time and uses O(n) space. This complexity is efficient because we only traverse the array a few times to build the left and right arrays.

What is the brute force solution for Count the Number of Incremovable Subarrays II?

The brute force approach for Count the Number of Incremovable Subarrays II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible subarray to see if removing it results in a strictly increasing array. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count the Number of Incremovable Subarrays II?

Count the Number of Incremovable Subarrays II uses the following concepts: Array, Two Pointers, Binary Search. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Count the Number of Incremovable Subarrays II?

Clarify the definition of incremovable subarrays. Discuss your thought process and potential edge cases with the interviewer. Practice explaining your approach clearly and concisely.

What are common mistakes when solving Count the Number of Incremovable Subarrays II?

Not checking edge cases like arrays of length 1. Overlooking the need to subtract n when counting total subarrays.

#2972

Count the Number of Incremovable Subarrays II

Hard

ArrayTwo PointersBinary SearchTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution leverages the idea of finding the longest strictly increasing prefix and suffix. By identifying these boundaries, we can efficiently count the valid incremovable subarrays without checking each one individually.

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Algorithm

3 steps

1Step 1: Identify the longest strictly increasing prefix from the start of the array.
2Step 2: Identify the longest strictly increasing suffix from the end of the array.
3Step 3: Count all possible subarrays that can be formed between these two boundaries.

solution.py12 lines

1def count_incremovable_subarrays(nums):
2    n = len(nums)
3    left = [0] * n
4    right = [0] * n
5    left[0] = 1
6    right[n - 1] = 1
7    for i in range(1, n):
8        left[i] = left[i - 1] + 1 if nums[i] > nums[i - 1] else 1
9    for i in range(n - 2, -1, -1):
10        right[i] = right[i + 1] + 1 if nums[i] < nums[i + 1] else 1
11    total = sum(left) + sum(right) - n
12    return total

ℹ

Complexity note: This complexity is efficient because we only traverse the array a few times to build the left and right arrays.

1Identifying boundaries of strictly increasing sequences is crucial.
2Counting valid subarrays can be optimized using prefix and suffix arrays.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.