Problem Description

You are given a circular array nums. In a circular array, the elements are arranged in a circle, meaning the first element and the last element are considered adjacent to each other.

Your task is to find the maximum absolute difference between any two adjacent elements in this circular array.

For example, if you have an array [1, 5, 3, 9]:

• The adjacent pairs are: (1,5), (5,3), (3,9), and (9,1) (note the last pair wraps around)
• The absolute differences are: |1-5| = 4, |5-3| = 2, |3-9| = 6, |9-1| = 8
• The maximum absolute difference would be 8

The solution uses Python's pairwise function with a clever trick: it appends the first element to the end of the array (nums + [nums[0]]) to handle the circular nature. This creates all adjacent pairs including the wrap-around pair. Then it calculates the absolute difference for each pair and returns the maximum value.

Intuition

The key insight is recognizing that in a circular array, we need to check all adjacent pairs, including the pair formed by the last and first elements.

When we think about finding the maximum difference between adjacent elements in a regular array, we would simply iterate through the array and compare each element with its next neighbor. However, the circular nature adds one extra comparison - the last element with the first element.

Instead of writing special logic to handle this wrap-around case separately, we can transform the problem into a simpler one. By appending the first element to the end of the array (nums + [nums[0]]), we effectively "unfold" the circle into a linear sequence where all adjacent relationships are preserved, including the circular connection.

For instance, if we have [1, 5, 3, 9], by appending the first element we get [1, 5, 3, 9, 1]. Now when we iterate through pairs:

• (1, 5), (5, 3), (3, 9), (9, 1)

We naturally capture all adjacent pairs including the wrap-around without any special handling. We can then simply calculate abs(a - b) for each pair and take the maximum. This elegant transformation reduces the circular array problem to a standard linear array problem, making the solution both simple and efficient.

Solution Approach

The solution follows a simulation approach where we traverse through all adjacent pairs in the circular array and track the maximum absolute difference.

The implementation uses a concise one-liner that combines several Python features:

1. Array Extension: We create nums + [nums[0]] to append the first element to the end of the array. This transforms the circular array problem into a linear one where all adjacent pairs can be accessed sequentially.

2. Pairwise Iteration: The pairwise() function from Python's itertools (available in Python 3.10+) generates consecutive pairs from the extended array. For an array [1, 5, 3, 9, 1], it produces pairs: (1, 5), (5, 3), (3, 9), (9, 1).

3. Absolute Difference Calculation: For each pair (a, b), we calculate abs(a - b) to get the absolute difference between adjacent elements.

4. Maximum Selection: The max() function is applied to the generator expression to find the largest absolute difference among all adjacent pairs.

The algorithm processes each element exactly once (plus one extra comparison for the wrap-around), making it run in O(n) time complexity where n is the length of the array. The space complexity is O(1) if we don't count the space used by the generator, or O(n) if we consider the temporary extended array.

This approach elegantly handles the circular nature of the array without requiring conditional logic or special cases, making the code both readable and efficient.

Example Walkthrough

Let's walk through the solution with a small example: nums = [4, 2, 8, 5]

Step 1: Identify all adjacent pairs in the circular array

• In a circular array, we have these adjacent pairs:
  • (4, 2) - first and second elements
  • (2, 8) - second and third elements
  • (8, 5) - third and fourth elements
  • (5, 4) - fourth and first elements (wrap-around)

Step 2: Transform the circular array

• To handle the wrap-around naturally, append the first element to the end:
• nums + [nums[0]] = [4, 2, 8, 5] + [4] = [4, 2, 8, 5, 4]

Step 3: Generate pairs from the extended array

• Using pairwise() on [4, 2, 8, 5, 4] gives us:
  • (4, 2)
  • (2, 8)
  • (8, 5)
  • (5, 4)

Step 4: Calculate absolute differences

• For each pair (a, b), compute abs(a - b):
  • abs(4 - 2) = 2
  • abs(2 - 8) = 6
  • abs(8 - 5) = 3
  • abs(5 - 4) = 1

Step 5: Find the maximum

• The maximum among [2, 6, 3, 1] is 6

Therefore, the maximum absolute difference between adjacent elements in this circular array is 6.

The beauty of this approach is that by simply appending the first element to the end, we convert the circular problem into a linear one, allowing us to use standard iteration techniques without any special logic for the wrap-around case.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the array nums.

The code creates a new list by concatenating nums with [nums[0]], which takes O(n) time. The pairwise function iterates through this list once to generate consecutive pairs, resulting in n pairs. For each pair (a, b), the code computes abs(a - b) in O(1) time. The max function then iterates through all n differences to find the maximum value, which also takes O(n) time. Therefore, the overall time complexity is O(n) + O(n) + O(n) = O(n).

Space Complexity: O(1) (or O(n) depending on interpretation).

The strict space complexity is O(n) because nums + [nums[0]] creates a new list of size n + 1. However, the reference answer states O(1), which likely refers to the auxiliary space complexity (excluding the space used for the input and temporary iterations). The pairwise function returns a generator that yields pairs on-the-fly without storing all pairs in memory simultaneously. The generator expression inside max also doesn't create an intermediate list. If we consider only the extra space beyond the input, the space complexity can be viewed as O(1) since we're only storing individual pairs and differences at any given time during iteration.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Python Version Compatibility Issue

The pairwise() function is only available in Python 3.10+. If you're using an older version of Python or submitting to a platform that doesn't support Python 3.10+, the code will fail with an ImportError.

Solution: Implement your own pairwise logic or use alternative approaches:

# Alternative 1: Manual iteration without pairwise
class Solution:
    def maxAdjacentDistance(self, nums: List[int]) -> int:
        n = len(nums)
        max_diff = 0
      
        for i in range(n):
            # Use modulo to handle circular indexing
            next_idx = (i + 1) % n
            max_diff = max(max_diff, abs(nums[i] - nums[next_idx]))
      
        return max_diff

# Alternative 2: Using zip instead of pairwise
class Solution:
    def maxAdjacentDistance(self, nums: List[int]) -> int:
        circular_nums = nums + [nums[0]]
        return max(abs(a - b) for a, b in zip(circular_nums, circular_nums[1:]))

2. Empty Array Handling

The current solution doesn't handle edge cases where the input array is empty or has only one element. Accessing nums[0] on an empty array will raise an IndexError.

Solution: Add validation for edge cases:

class Solution:
    def maxAdjacentDistance(self, nums: List[int]) -> int:
        if len(nums) <= 1:
            return 0  # No adjacent pairs exist
      
        circular_nums = nums + [nums[0]]
        return max(abs(a - b) for a, b in pairwise(circular_nums))

3. Memory Overhead for Large Arrays

Creating nums + [nums[0]] creates a new list that copies all elements, which can be inefficient for very large arrays.

Solution: Use an index-based approach that doesn't create a new array:

class Solution:
    def maxAdjacentDistance(self, nums: List[int]) -> int:
        if len(nums) <= 1:
            return 0
      
        max_diff = abs(nums[-1] - nums[0])  # Handle wrap-around case
      
        for i in range(len(nums) - 1):
            max_diff = max(max_diff, abs(nums[i] - nums[i + 1]))
      
        return max_diff

4. Assuming Non-Empty Generator

When using max() on a generator expression, if the generator is empty (which could happen with an empty input), it will raise a ValueError.

Solution: Provide a default value to max():

class Solution:
    def maxAdjacentDistance(self, nums: List[int]) -> int:
        if not nums:
            return 0
          
        circular_nums = nums + [nums[0]]
        return max((abs(a - b) for a, b in pairwise(circular_nums)), default=0)