What is the time complexity of Longest Subarray of 1's After Deleting One Element?

The optimal solution for Longest Subarray of 1's After Deleting One Element runs in O(n) time and uses O(1) space. The time complexity is O(n) because we traverse the array with two pointers, each moving at most n times, leading to a linear scan.

What is the brute force solution for Longest Subarray of 1's After Deleting One Element?

The brute force approach for Longest Subarray of 1's After Deleting One Element is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray after removing one element. This means we will iterate through the array, remove each element one at a time, and then find the longest contiguous subarray of 1's. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Subarray of 1's After Deleting One Element?

Longest Subarray of 1's After Deleting One Element uses the following concepts: Array, Dynamic Programming, Sliding Window. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Longest Subarray of 1's After Deleting One Element?

Explain your thought process clearly while coding. Consider edge cases, such as arrays with only 1's or only 0's. Practice sliding window problems to become comfortable with the technique.

What are common mistakes when solving Longest Subarray of 1's After Deleting One Element?

Not properly managing the count of zeros in the sliding window. Forgetting to account for the case where the entire array is 1's.

#1493

Longest Subarray of 1's After Deleting One Element

Medium

ArrayDynamic ProgrammingSliding WindowSliding WindowTwo Pointers

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to maintain a window that contains at most one zero. This way, we can efficiently find the longest subarray of 1's after deleting one element.

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Algorithm

5 steps

1Step 1: Initialize two pointers (left and right) and a variable to count zeros.
2Step 2: Expand the right pointer to include elements in the window until we encounter more than one zero.
3Step 3: If the count of zeros exceeds one, move the left pointer to shrink the window until we have at most one zero.
4Step 4: Calculate the length of the current valid window and update the maximum length.
5Step 5: Return the maximum length found.

solution.py13 lines

1def longestSubarray(nums):
2    left = 0
3    max_length = 0
4    zero_count = 0
5    for right in range(len(nums)):
6        if nums[right] == 0:
7            zero_count += 1
8        while zero_count > 1:
9            if nums[left] == 0:
10                zero_count -= 1
11            left += 1
12        max_length = max(max_length, right - left)
13    return max_length

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Complexity note: The time complexity is O(n) because we traverse the array with two pointers, each moving at most n times, leading to a linear scan.

1Using a sliding window allows us to efficiently manage the count of zeros.
2We can find the longest subarray by adjusting the window size dynamically.

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