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Maximum Difference Between Adjacent Elements in a Circular Array

Easy

Asked by:

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Topics:

Arrays

Given a circular array nums, find the maximum absolute difference between adjacent elements.

Note: In a circular array, the first and last elements are adjacent.

Example 1:

Input: nums = [1,2,4]

Output: 3

Explanation:

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

Explanation:

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

Maximum Difference Between Adjacent Elements in a Circular Array

Easy

Asked by:

44 views

Topics:

Arrays

Given a circular array nums, find the maximum absolute difference between adjacent elements.

Note: In a circular array, the first and last elements are adjacent.

Example 1:

Input: nums = [1,2,4]

Output: 3

Explanation:

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

Explanation:

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

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of integer values within the array? Can the array contain negative numbers, zeros, or only positive numbers?
What should I return if the input array is null, empty, or contains only one element? Is there a default return value?
Are duplicate values allowed in the array, and if so, should I consider all possible adjacent pairs when calculating the maximum difference?
By 'adjacent elements in a circular array', do you mean that the last and first elements are also considered adjacent?
Is there an integer overflow risk when calculating the difference between two adjacent numbers? Should I consider using a larger data type for intermediate calculations?

Brute Force Solution

Approach

We need to find the biggest difference between any two neighboring numbers when the numbers are arranged in a circle. The brute force way is to simply check every possible pair of neighboring numbers and compare their differences.

Here's how the algorithm would work step-by-step:

Consider the first number in the circle and find the difference between it and the number next to it.
Remember this difference.
Now, do the same thing for the second number and the number next to it in the circle.
If this new difference is bigger than the one we remembered, replace the remembered difference with this new one.
Keep doing this for every number in the circle, always comparing the difference between a number and its neighbor with what we've remembered.
Because it's a circle, also check the difference between the very last number and the very first number.
After checking every pair of neighbors, the difference we've remembered is the biggest difference between any two neighbors in the circular arrangement.

Code Implementation

def find_max_circular_difference_brute_force(circular_array):
    array_length = len(circular_array)

    # Initialize the maximum difference to a small value.
    maximum_difference = 0

    for index in range(array_length):

        # Calculate index of next element
        next_index = (index + 1) % array_length

        # Find the difference between adjacent
        difference = abs(circular_array[index] - circular_array[next_index])

        # Update maximum difference if necessary
        if difference > maximum_difference:
            maximum_difference = difference

    return maximum_difference

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each of the n elements in the array once. For each element, it calculates the difference with its adjacent element in the circular array, which takes constant time, O(1). The maximum difference is updated in each iteration. Since the number of iterations is directly proportional to the size of the input array, the overall time complexity is O(n).

Space Complexity

O(1) – The provided plain English explanation only describes storing the current maximum difference found so far. This requires a single variable to hold this maximum difference. No other data structures are used or created that scale with the input size N (the number of elements in the circular array). Therefore, the space complexity is constant.

Optimal Solution

Approach

The key to efficiently solving this problem involves understanding that the largest difference will always occur between the biggest and smallest numbers in the input. The approach focuses on finding these two values and then cleverly considering the circular nature of the list.

Here's how the algorithm would work step-by-step:

Find the largest and smallest numbers in the given list.
Calculate the difference between these two numbers; this is a potential candidate for the maximum difference.
Also, calculate the difference between the first and last elements of the list. Treat this list as a circle, so the end connects to the beginning.
The answer is the bigger of those two differences.

Code Implementation

def max_circular_difference(number_list):
    list_length = len(number_list)

    if list_length < 2:
        return 0

    maximum_element = number_list[0]
    minimum_element = number_list[0]

    for element in number_list:
        maximum_element = max(maximum_element, element)
        minimum_element = min(minimum_element, element)

    # Calculate difference between max and min
    max_min_difference = maximum_element - minimum_element

    # Account for circularity, diff between first and last.
    circular_difference = abs(number_list[0] - number_list[-1])

    #Return the greater of the two differences
    return max(max_min_difference, circular_difference)

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The solution finds the largest and smallest numbers in the list by iterating through it once. This requires examining each of the n elements of the input list. Calculating the differences between the largest and smallest element and the first and last elements are constant time operations. Therefore, the dominant operation is the single pass through the list, resulting in a time complexity of O(n).

Space Complexity

O(1) – The algorithm only requires storing the largest and smallest numbers found in the input list. These are stored in variables, and the number of such variables is constant regardless of the input list's size (N). Additionally, calculating differences does not require extra space, resulting in a space complexity that remains constant. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty or null array	Return 0, or throw an IllegalArgumentException, or return a predefined error code depending on requirements, since there's no difference to calculate.
Array with only one element	Return 0 since the difference between an element and itself is zero (circular).
Array with two elements	Return the absolute difference between the two elements directly as that is the only adjacent pair.
Array with all elements being the same value	Return 0 as the difference between adjacent elements will always be zero.
Array with large positive and negative numbers (potential overflow)	Use a data type that can hold larger values (e.g., long) to avoid potential integer overflow when calculating the difference.
Array with a large number of elements (performance concern)	An O(n) solution is generally acceptable; ensure the code doesn't inadvertently introduce higher complexity (e.g., nested loops).
Array with only negative numbers	The algorithm should correctly calculate the absolute difference even with all negative numbers.
Array with numbers near Integer.MAX_VALUE and Integer.MIN_VALUE	Use `long` to prevent integer overflow when calculating the absolute difference of adjacent elements.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of integer values within the array? Can the array contain negative numbers, zeros, or only positive numbers?
What should I return if the input array is null, empty, or contains only one element? Is there a default return value?
Are duplicate values allowed in the array, and if so, should I consider all possible adjacent pairs when calculating the maximum difference?
By 'adjacent elements in a circular array', do you mean that the last and first elements are also considered adjacent?
Is there an integer overflow risk when calculating the difference between two adjacent numbers? Should I consider using a larger data type for intermediate calculations?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider the first number in the circle and find the difference between it and the number next to it.
Remember this difference.
Now, do the same thing for the second number and the number next to it in the circle.
If this new difference is bigger than the one we remembered, replace the remembered difference with this new one.
Keep doing this for every number in the circle, always comparing the difference between a number and its neighbor with what we've remembered.
Because it's a circle, also check the difference between the very last number and the very first number.
After checking every pair of neighbors, the difference we've remembered is the biggest difference between any two neighbors in the circular arrangement.

Code Implementation

def find_max_circular_difference_brute_force(circular_array):
    array_length = len(circular_array)

    # Initialize the maximum difference to a small value.
    maximum_difference = 0

    for index in range(array_length):

        # Calculate index of next element
        next_index = (index + 1) % array_length

        # Find the difference between adjacent
        difference = abs(circular_array[index] - circular_array[next_index])

        # Update maximum difference if necessary
        if difference > maximum_difference:
            maximum_difference = difference

    return maximum_difference

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Find the largest and smallest numbers in the given list.
Calculate the difference between these two numbers; this is a potential candidate for the maximum difference.
Also, calculate the difference between the first and last elements of the list. Treat this list as a circle, so the end connects to the beginning.
The answer is the bigger of those two differences.

Code Implementation

def max_circular_difference(number_list):
    list_length = len(number_list)

    if list_length < 2:
        return 0

    maximum_element = number_list[0]
    minimum_element = number_list[0]

    for element in number_list:
        maximum_element = max(maximum_element, element)
        minimum_element = min(minimum_element, element)

    # Calculate difference between max and min
    max_min_difference = maximum_element - minimum_element

    # Account for circularity, diff between first and last.
    circular_difference = abs(number_list[0] - number_list[-1])

    #Return the greater of the two differences
    return max(max_min_difference, circular_difference)

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty or null array	Return 0, or throw an IllegalArgumentException, or return a predefined error code depending on requirements, since there's no difference to calculate.
Array with only one element	Return 0 since the difference between an element and itself is zero (circular).
Array with two elements	Return the absolute difference between the two elements directly as that is the only adjacent pair.
Array with all elements being the same value	Return 0 as the difference between adjacent elements will always be zero.
Array with large positive and negative numbers (potential overflow)	Use a data type that can hold larger values (e.g., long) to avoid potential integer overflow when calculating the difference.
Array with a large number of elements (performance concern)	An O(n) solution is generally acceptable; ensure the code doesn't inadvertently introduce higher complexity (e.g., nested loops).
Array with only negative numbers	The algorithm should correctly calculate the absolute difference even with all negative numbers.
Array with numbers near Integer.MAX_VALUE and Integer.MIN_VALUE	Use `long` to prevent integer overflow when calculating the absolute difference of adjacent elements.