algoadvance

Given an integer array nums and an integer threshold, the task is to find the length of the longest subarray that meets the following criteria:

Clarifying Questions

  1. What are the expected inputs?
    • The input nums is a list of integers.
    • The input threshold is a single integer.
  2. What should be returned?
    • The length of the longest subarray that fulfills the given criteria.
  3. Can the array contain negative numbers?
    • It depends on the specific problem constraints, but typically arrays can contain negative values unless otherwise stated.
  4. What if multiple subarrays have the same length?
    • We only need to return the length, so it doesn’t matter if there are multiple subarrays of the same maximum length.
  5. Is the array guaranteed to contain at least one element?
    • This will be assumed unless specified otherwise in the problem.
  6. Can the threshold be negative?
    • Typically, in problems like this, the threshold is non-negative, but both positive and negative numbers below the threshold should be considered.

Strategy

  1. Initialize Variables:
    • Use two pointers to track the start and end of the current valid subarray.
    • Keep a variable to store the length of the longest valid subarray found.
  2. Iterate over the Array:
    • Check each element and its adjacent element to ensure they’re alternating and below the threshold.
    • Maintain a running count of the current valid subarray length.
    • Update the maximum length whenever a longer valid subarray is found.
  3. Edge Cases:
    • Handle empty arrays by returning 0.
    • Handle arrays where none or only some elements meet the criteria.

Code

def longest_even_odd_subarray(nums, threshold):
    max_length = 0
    current_length = 0
    n = len(nums)
    
    if n == 0:
        return 0
    
    for i in range(n):
        if nums[i] > threshold:
            current_length = 0
            continue
        
        if i == 0 or (nums[i] <= threshold and nums[i-1] <= threshold and 
                      ((nums[i] % 2 == 0 and nums[i-1] % 2 != 0) or (nums[i] % 2 != 0 and nums[i-1] % 2 == 0))):
            current_length += 1
        else:
            current_length = 1
        
        max_length = max(max_length, current_length)
    
    return max_length

# Example usage:
nums = [3, 2, 5, 4, 5, 6, 7, 8]
threshold = 5
print(longest_even_odd_subarray(nums, threshold))  # Output: 4

Time Complexity

The time complexity of this approach is O(n) where n is the number of elements in the array. This is because we are making a single pass through the array to determine the longest valid subarray length.

Space Complexity

The space complexity is O(1) because we only use a few extra variables for tracking lengths and pointers, independent of the input size.

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