How do you solve Longest Even Odd Subarray With Threshold on LeetCode?

Longest Even Odd Subarray With Threshold (LeetCode #2760) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The subarray must start with an even number and alternate between even and odd.

What is the brute force solution for Longest Even Odd Subarray With Threshold?

The brute force approach for Longest Even Odd Subarray With Threshold is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray to see if it meets the conditions. This is straightforward but inefficient, as it examines all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Even Odd Subarray With Threshold?

Longest Even Odd Subarray With Threshold uses the following concepts: Array, Sliding Window. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Longest Even Odd Subarray With Threshold?

Clarify the problem requirements before jumping into coding. Consider edge cases, such as arrays with only one element or all elements exceeding the threshold. Explain your thought process while coding to demonstrate your understanding.

What are common mistakes when solving Longest Even Odd Subarray With Threshold?

Not checking if the starting element is even before proceeding. Failing to properly alternate between even and odd numbers in the subarray.

#2760

Longest Even Odd Subarray With Threshold

Easy

ArraySliding WindowSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a sliding window approach to efficiently find the longest valid subarray without needing to check every possible combination.

⚙️

Algorithm

4 steps

1Step 1: Initialize pointers for the start of the window and a variable for the maximum length.
2Step 2: Expand the window by moving the end pointer and check conditions for validity.
3Step 3: If the subarray becomes invalid, move the start pointer to the right until it becomes valid again.
4Step 4: Update the maximum length whenever a valid subarray is found.

solution.py14 lines

1def longest_even_odd_subarray(nums, threshold):
2    max_length = 0
3    n = len(nums)
4    start = 0
5    while start < n:
6        if nums[start] % 2 == 0:
7            end = start
8            current_length = 1
9            while end + 1 < n and nums[end + 1] <= threshold and nums[end + 1] % 2 != nums[end] % 2:
10                end += 1
11                current_length += 1
12            max_length = max(max_length, current_length)
13        start += 1
14    return max_length

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once with the sliding window technique. The space complexity is O(1) since we are using a constant amount of extra space.

1The subarray must start with an even number and alternate between even and odd.
2All elements in the subarray must be less than or equal to the threshold.

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