#3423

Easy

auto_awesomeArray-driven solution strategy

LeetCode Problem Workspace

Maximum Difference Between Adjacent Elements in a Circular Array

Find the maximum absolute difference between adjacent elements in a circular array using a straightforward array-driven approach.

array

Problem Statement

Given a circular array nums of integers, determine the maximum absolute difference between any two adjacent elements. Remember that the first and last elements are considered adjacent because of the circular structure.

You are provided an array nums with at least two elements and integer values between -100 and 100. Return the largest absolute difference found between adjacent pairs, ensuring you include the circular connection from the last element to the first.

Examples

Example 1

Input: nums = [1,2,4]

Output: 3

Because nums is circular, nums[0] and nums[2] are adjacent. They have the maximum absolute difference of |4 - 1| = 3 .

Example 2

Input: nums = [-5,-10,-5]

Output: 5

The adjacent elements nums[0] and nums[1] have the maximum absolute difference of |-5 - (-10)| = 5 .

Constraints

2 <= nums.length <= 100
-100 <= nums[i] <= 100

Solution Approach

Single Pass Traversal

Iterate from the first element to the last, computing the absolute difference between each element and its next neighbor. Include the difference between the last and first elements to account for the circular array property.

Tracking Maximum Difference

Maintain a variable to store the maximum difference found so far. Update this variable whenever a larger difference is encountered during the traversal.

Constant Space Efficiency

No extra data structures are needed beyond variables for the current difference and maximum difference. The solution runs in O(n) time and O(1) space, suitable for the given constraints.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

Time complexity is O(n) because we traverse the array once, comparing each adjacent pair. Space complexity is O(1) since we only store variables for the maximum and current differences.

What Interviewers Usually Probe

The candidate understands array traversal and circular adjacency handling.
They correctly compute absolute differences and maintain a running maximum.
They can explain why extra data structures are unnecessary for optimal space.

Common Pitfalls or Variants

Common pitfalls

Forgetting to include the difference between the last and first elements.
Using additional arrays unnecessarily, increasing space complexity.
Miscomputing absolute difference, leading to incorrect maximum values.

Follow-up variants

Finding the minimum absolute difference between adjacent elements in a circular array.
Computing the maximum difference in a rotated sorted array.
Handling arrays with floating-point numbers instead of integers.

FAQ

What is the maximum difference between adjacent elements in a circular array?

It is the largest absolute difference found between consecutive elements, including the difference between the last and first elements due to the circular structure.

Do I need extra space to solve this problem?

No, you can compute the maximum difference using constant space by tracking only the current and maximum differences.

Can negative numbers affect the maximum difference?

Yes, absolute difference calculation accounts for negative values, so the maximum difference is still correctly identified.

Is the solution pattern always a single pass through the array?

Yes, traversing once while comparing adjacent elements, including the circular connection, efficiently finds the answer.

What array-driven strategies are key for this problem?

Iterate through each element to its neighbor, compute absolute differences, and maintain the maximum, while remembering the circular adjacency of the first and last elements.

terminal

Solution

Solution 1: Simulation

We traverse the array $\textit{nums}$, calculate the absolute difference between adjacent elements, and maintain the maximum absolute difference. Finally, we compare it with the absolute difference between the first and last elements and take the maximum value.

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class Solution:
    def maxAdjacentDistance(self, nums: List[int]) -> int:
        return max(abs(a - b) for a, b in pairwise(nums + [nums[0]]))

Continue Practicing

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