What is the time complexity of Longest Non-Decreasing Subarray After Replacing at Most One Element?

The optimal solution for Longest Non-Decreasing Subarray After Replacing at Most One Element runs in O(n) time and uses O(n) space. This solution runs in linear time due to single passes for prefix and suffix arrays, leading to efficient performance.

What is the brute force solution for Longest Non-Decreasing Subarray After Replacing at Most One Element?

The brute force approach for Longest Non-Decreasing Subarray After Replacing at Most One Element is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible subarray by replacing each element with all possible values. This brute force method ensures we find the longest non-decreasing subarray but is inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Non-Decreasing Subarray After Replacing at Most One Element?

Longest Non-Decreasing Subarray After Replacing at Most One Element uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: .

What are the interview tips for Longest Non-Decreasing Subarray After Replacing at Most One Element?

Clarify constraints and edge cases before coding. Discuss your thought process and approach with the interviewer. Practice similar problems to build confidence.

What are common mistakes when solving Longest Non-Decreasing Subarray After Replacing at Most One Element?

Not considering edge cases with single elements. Overlooking the need to check adjacent elements for valid replacements.

#3738

Longest Non-Decreasing Subarray After Replacing at Most One Element

Medium

ArrayDynamic Programming

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)

We can use two pointers to find the longest non-decreasing subarray while allowing one replacement. This approach efficiently checks the conditions without redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Create two arrays, prefix and suffix, to store the lengths of non-decreasing subarrays ending and starting at each index.
2Step 2: Iterate through the array to fill the prefix and suffix arrays.
3Step 3: Calculate the maximum length by checking the combination of prefix and suffix lengths around each index.

solution.py14 lines

1def longestNonDecreasing(nums):
2    n = len(nums)
3    if n == 1: return 1
4    prefix = [1] * n
5    suffix = [1] * n
6    for i in range(1, n):
7        if nums[i] >= nums[i - 1]: prefix[i] = prefix[i - 1] + 1
8    for i in range(n - 2, -1, -1):
9        if nums[i] <= nums[i + 1]: suffix[i] = suffix[i + 1] + 1
10    max_length = max(prefix)
11    for i in range(1, n - 1):
12        if nums[i - 1] <= nums[i + 1]:
13            max_length = max(max_length, prefix[i - 1] + suffix[i + 1])
14    return max_length

ℹ

Complexity note: This solution runs in linear time due to single passes for prefix and suffix arrays, leading to efficient performance.

1Replacing an element can connect two non-decreasing segments.
2Maintaining prefix and suffix arrays allows efficient length calculation.

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