Array With Elements Not Equal to Average of Neighbors

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ArraysGreedy Algorithms

You are given a 0-indexed array nums of distinct integers. You want to rearrange the elements in the array such that every element in the rearranged array is not equal to the average of its neighbors.

More formally, the rearranged array should have the property such that for every i in the range 1 <= i < nums.length - 1, (nums[i-1] + nums[i+1]) / 2 is not equal to nums[i].

Return any rearrangement of nums that meets the requirements.

For example:

nums = [1, 2, 3, 4, 5] A valid output might be [1, 2, 4, 5, 3] because:

When i=1, nums[i] = 2, and the average of its neighbors is (1+4) / 2 = 2.5.
When i=2, nums[i] = 4, and the average of its neighbors is (2+5) / 2 = 3.5.
When i=3, nums[i] = 5, and the average of its neighbors is (4+3) / 2 = 3.5.

As another example, consider:

nums = [6, 2, 0, 9, 7] A valid output might be [9, 7, 6, 2, 0] because:

When i=1, nums[i] = 7, and the average of its neighbors is (9+6) / 2 = 7.5.
When i=2, nums[i] = 6, and the average of its neighbors is (7+2) / 2 = 4.5.
When i=3, nums[i] = 2, and the average of its neighbors is (6+0) / 2 = 3.

How would you implement an algorithm to solve this problem efficiently?

Array With Elements Not Equal to Average of Neighbors

Medium

Uber

1 view

Topics:

ArraysGreedy Algorithms

You are given a 0-indexed array nums of distinct integers. You want to rearrange the elements in the array such that every element in the rearranged array is not equal to the average of its neighbors.

More formally, the rearranged array should have the property such that for every i in the range 1 <= i < nums.length - 1, (nums[i-1] + nums[i+1]) / 2 is not equal to nums[i].

Return any rearrangement of nums that meets the requirements.

For example:

nums = [1, 2, 3, 4, 5] A valid output might be [1, 2, 4, 5, 3] because:

When i=1, nums[i] = 2, and the average of its neighbors is (1+4) / 2 = 2.5.
When i=2, nums[i] = 4, and the average of its neighbors is (2+5) / 2 = 3.5.
When i=3, nums[i] = 5, and the average of its neighbors is (4+3) / 2 = 3.5.

As another example, consider:

nums = [6, 2, 0, 9, 7] A valid output might be [9, 7, 6, 2, 0] because:

When i=1, nums[i] = 7, and the average of its neighbors is (9+6) / 2 = 7.5.
When i=2, nums[i] = 6, and the average of its neighbors is (7+2) / 2 = 4.5.
When i=3, nums[i] = 2, and the average of its neighbors is (6+0) / 2 = 3.

How would you implement an algorithm to solve this problem efficiently?

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 input array?
Can the input array be empty or contain only one or two elements?
If multiple valid arrangements exist, is any valid arrangement acceptable, or is there a preferred one?
Does the input array contain only integers or could it contain floats or other data types?
Are duplicate numbers allowed in the input array, and if so, how should they be handled?

Brute Force Solution

Approach

The brute force approach for this problem is like trying every possible way to rearrange the numbers. We want to find an arrangement where no number is exactly the average of its immediate neighbors. We essentially explore all arrangements and check each one to see if it meets the criteria.

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

Start with the original order of the numbers.
Generate the very next possible order of the numbers.
Check if, in this new order, each number is NOT equal to the average of the numbers directly next to it.
If every number passes this check, then we've found a valid arrangement, and we can stop.
If not, we discard this arrangement and generate the next possible order.
We keep repeating steps 2-5 until we find a valid arrangement or we've tried absolutely every possible order of the numbers.

Code Implementation

def find_arrangement_brute_force(numbers):
    import itertools

    # Generate all possible permutations of the numbers
    for permutation in itertools.permutations(numbers):
        number_list = list(permutation)
        is_valid = True

        # Check if the current permutation is a valid arrangement
        for index in range(len(number_list)): 

            # Skip the check for the first and last elements
            if index == 0 or index == len(number_list) - 1:
                continue

            # Check if the current number is the average of its neighbors
            if number_list[index] == (number_list[index - 1] + number_list[index + 1]) / 2:
                is_valid = False
                break

        # If the arrangement is valid, return it
        if is_valid:
            return number_list

    return []

Big(O) Analysis

Time Complexity

O(n! * n) – The provided brute force approach involves generating all possible permutations of the input array of size n. Generating all permutations takes O(n!) time. For each permutation, we need to check if any element is equal to the average of its neighbors. This check involves iterating through the array once, taking O(n) time. Therefore, the overall time complexity is O(n! * n).

Space Complexity

O(N) – The brute force approach involves generating permutations of the input array. A common way to generate permutations requires storing the current permutation, which is a copy of the input array. This means we need an auxiliary array of size N, where N is the number of elements in the input array, to store the rearranged numbers during each permutation check. Therefore, the space complexity is O(N).

Optimal Solution

Approach

The key is to realize that by sorting the numbers, we can easily arrange them to satisfy the condition. We place smaller and larger numbers alternately to ensure that no number equals the average of its neighbors.

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

First, put all the numbers in order from smallest to largest.
Next, create a new list by alternating between taking a small number then a large number from the sorted numbers.
Place the smallest number in the first spot, then the largest number in the second spot, the second smallest in the third, and the second largest in the fourth spot, and so on.
This guarantees that each number is different from the average of its two neighbors because a smaller number will always be next to a larger number.

Code Implementation

def rearrange_array(numbers):
    numbers.sort()

    rearranged_numbers = [0] * len(numbers)
    left_index = 0
    right_index = len(numbers) - 1

    # Place elements alternately from the start and end.
    for i in range(len(numbers)):
        if i % 2 == 0:

            rearranged_numbers[i] = numbers[left_index]
            left_index += 1

        else:
            # Arrange larger number on even indices
            rearranged_numbers[i] = numbers[right_index]
            right_index -= 1

    return rearranged_numbers

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this solution is sorting the input array of size n. Efficient sorting algorithms, such as merge sort or quicksort, typically have a time complexity of O(n log n). The subsequent rearrangement of elements involves iterating through the sorted array once, which takes O(n) time. Since O(n log n) grows faster than O(n), the overall time complexity of the algorithm is determined by the sorting step. Therefore, the time complexity is O(n log n).

Space Complexity

O(N) – The provided solution creates a new list to store the rearranged elements. The size of this new list is equal to the number of elements in the input array, which we denote as N. Therefore, the auxiliary space required is proportional to N. Consequently, the space complexity of the solution is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty array immediately as no rearrangement is possible.
Array with only one or two elements	For arrays of size one, return the array as is; arrays of size two can be swapped if identical, or returned as is.
Array with all identical elements	A valid arrangement is impossible; return the original array if the prompt allows or handle as an error if a solution is required.
Array with a highly skewed distribution of values (e.g., one extremely large value)	Sorting the array ensures the extreme value is placed to satisfy the neighbor condition.
Array containing negative numbers, zeros, and positive numbers	The sorting approach is resilient to negative numbers and zeros, placing them appropriately.
Large input array (e.g., size 10^5) that could cause performance issues	Use an efficient sorting algorithm like merge sort or quicksort which has O(n log n) time complexity.
Integer overflow during calculations with large numbers in the array.	Check for potential overflows when calculating the average of neighbors if using integer arithmetic, and consider using larger data types if needed.
Multiple possible valid solutions exist; ensure algorithm finds any valid one.	The specific arrangement produced by the sorting and swapping approach is just one of the possible valid arrangements.