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

Count the Number of Incremovable Subarrays I (LeetCode #2970) 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: Understanding the boundaries of increasing sequences helps in counting valid subarrays efficiently.

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

The brute force approach for Count the Number of Incremovable Subarrays I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible subarrays to see if removing them results in a strictly increasing array. This is straightforward but inefficient as it examines every possible combination. 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 I?

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

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

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases and how they might affect your solution. Explain your thought process while coding; it helps interviewers understand your approach.

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

Failing to check if the remaining array is strictly increasing after removing a subarray. Not considering edge cases, such as arrays that are already strictly increasing.

#2970

Count the Number of Incremovable Subarrays I

Easy

ArrayTwo PointersBinary SearchEnumerationTwo PointersSliding WindowArray

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 identifying the longest increasing prefix and suffix. By determining the boundaries of the incremovable subarrays, we can count them efficiently without checking every possibility.

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Algorithm

4 steps

1Step 1: Identify the longest increasing prefix of the array.
2Step 2: Identify the longest increasing suffix of the array.
3Step 3: Count the number of valid subarrays that can be formed between these two boundaries.
4Step 4: Return the total count of incremovable subarrays.

solution.py14 lines

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

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times. The space complexity is O(n) due to the additional arrays used to store the lengths of increasing sequences.

1Understanding the boundaries of increasing sequences helps in counting valid subarrays efficiently.
2Using auxiliary arrays can simplify the problem of checking conditions on subarrays.

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