What is the time complexity of Longest Alternating Subarray After Removing At Most One Element?

The optimal solution for Longest Alternating Subarray After Removing At Most One Element runs in O(n) time and uses O(n) space. Filling the left and right arrays takes linear time, leading to O(n) complexity.

What is the brute force solution for Longest Alternating Subarray After Removing At Most One Element?

The brute force approach for Longest Alternating Subarray After Removing At Most One Element is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible subarray and determine if it is alternating. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Alternating Subarray After Removing At Most One Element?

Longest Alternating Subarray After Removing At Most One Element uses the following concepts: Array, Dynamic Programming, Enumeration. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Longest Alternating Subarray After Removing At Most One Element?

Explain your thought process clearly. Consider edge cases like small arrays. Practice similar problems to build intuition.

What are common mistakes when solving Longest Alternating Subarray After Removing At Most One Element?

Not considering equal adjacent elements. Overlooking the impact of removing elements.

#3830

Longest Alternating Subarray After Removing At Most One Element

Hard

ArrayDynamic ProgrammingEnumerationDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Use dynamic programming to track lengths of alternating subarrays from both ends, allowing for one removal.

⚙️

Algorithm

3 steps

1Step 1: Create two arrays, left and right, to store lengths of alternating sequences ending and starting at each index.
2Step 2: Fill the left array from left to right and the right array from right to left.
3Step 3: Calculate the maximum length considering one removal by checking adjacent elements.

solution.py19 lines

1def longestAlternating(nums):
2    n = len(nums)
3    left = [1] * n
4    right = [1] * n
5    for i in range(1, n):
6        if (nums[i - 1] < nums[i]):
7            left[i] = left[i - 1] + 1
8        elif (nums[i - 1] > nums[i]):
9            left[i] = left[i - 1] + 1
10    for i in range(n - 2, -1, -1):
11        if (nums[i] < nums[i + 1]):
12            right[i] = right[i + 1] + 1
13        elif (nums[i] > nums[i + 1]):
14            right[i] = right[i + 1] + 1
15    max_len = max(left)
16    for i in range(1, n - 1):
17        if (nums[i - 1] < nums[i + 1]):
18            max_len = max(max_len, left[i - 1] + right[i + 1])
19    return max_len

ℹ

Complexity note: Filling the left and right arrays takes linear time, leading to O(n) complexity.

1Identifying alternating patterns is crucial.
2Removing one element can bridge two alternating sequences.

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