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Longest Even Odd Subarray With Threshold

Difficulty: Easy

Problem Description

You are given a 0-indexed integer array nums and an integer threshold. Find the length of the longest subarray of nums starting at index l and ending at index r (0 <= l <= r < nums.length) that satisfies the following conditions:

  • nums[l] % 2 == 0
  • For all indices i in the range [l, r - 1], nums[i] % 2 != nums[i + 1] % 2
  • For all indices i in the range [l, r], nums[i] <= threshold

Return an integer denoting the length of the longest such subarray.

Key Insights

  • The first element of the subarray must be even.
  • The subarray must alternate between even and odd numbers.
  • All elements in the subarray must not exceed the given threshold.
  • A sliding window approach can be used to efficiently find the longest valid subarray.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(1)

Solution

The solution involves using a sliding window technique. We maintain two pointers to represent the current window of the subarray. We begin by checking if the first element is even and less than or equal to the threshold. If it is, we proceed to expand the window to the right, checking the conditions for each subsequent element. If any condition fails (either that the element is not even/odd as required or exceeds the threshold), we adjust the left pointer to try and find a new valid subarray. The maximum length of any valid subarray found during this process is stored and returned.

Code Solutions

def longest_even_odd_subarray(nums, threshold):
    max_length = 0
    n = len(nums)
    l = 0
    
    while l < n:
        if nums[l] % 2 == 0 and nums[l] <= threshold:
            r = l
            current_length = 1
            
            while r + 1 < n and nums[r + 1] % 2 != nums[r] % 2 and nums[r + 1] <= threshold:
                current_length += 1
                r += 1
            
            max_length = max(max_length, current_length)
        
        l += 1
    
    return max_length
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