Wiggle Sort II

Medium

Amazon

2 views

Topics:

ArraysGreedy Algorithms

Let's explore a problem involving array manipulation. Imagine you're given an unsorted integer array, and your task is to reorder it in a specific "wiggle" pattern. The goal is to arrange the elements such that nums[0] < nums[1] > nums[2] < nums[3]... and so on. Essentially, elements at even indices should be smaller than their adjacent odd-indexed elements, and vice versa. You can assume that the input array always has a valid solution.

For instance, consider the array [1, 5, 1, 1, 6, 4]. A valid wiggle sort of this array would be [1, 6, 1, 5, 1, 4]. Another acceptable output would be [1, 4, 1, 5, 1, 6]. Both satisfy the condition.

Another example, the array [1, 3, 2, 2, 3, 1] has a wiggle sort of [2, 3, 1, 3, 1, 2].

Constraints to keep in mind:

The length of the array will be between 1 and 50,000.
Each element in the array will be a non-negative integer between 0 and 5000.
A valid wiggle sort is always possible for the given input.

Your task is to implement an algorithm to perform this wiggle sort efficiently. Can you do it in O(n) time and in-place with O(1) extra space?

Wiggle Sort II

Medium

Amazon

2 views

Topics:

ArraysGreedy Algorithms

Let's explore a problem involving array manipulation. Imagine you're given an unsorted integer array, and your task is to reorder it in a specific "wiggle" pattern. The goal is to arrange the elements such that nums[0] < nums[1] > nums[2] < nums[3]... and so on. Essentially, elements at even indices should be smaller than their adjacent odd-indexed elements, and vice versa. You can assume that the input array always has a valid solution.

For instance, consider the array [1, 5, 1, 1, 6, 4]. A valid wiggle sort of this array would be [1, 6, 1, 5, 1, 4]. Another acceptable output would be [1, 4, 1, 5, 1, 6]. Both satisfy the condition.

Another example, the array [1, 3, 2, 2, 3, 1] has a wiggle sort of [2, 3, 1, 3, 1, 2].

Constraints to keep in mind:

The length of the array will be between 1 and 50,000.
Each element in the array will be a non-negative integer between 0 and 5000.
A valid wiggle sort is always possible for the given input.

Your task is to implement an algorithm to perform this wiggle sort efficiently. Can you do it in O(n) time and in-place with O(1) extra space?

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 values within the input array? Can I expect negative numbers, zeros, or only positive integers?
Can the input array be empty or null? If so, what should I return or do?
If there are multiple possible wiggle sorted arrays, is any valid arrangement acceptable, or is there a specific wiggle sort order you'd prefer?
How should duplicate values be handled? Should they be placed strategically to satisfy the wiggle condition?
Is the input array guaranteed to have a wiggle sort solution, or is it possible that no wiggle sort exists for the given input array?

Brute Force Solution

Approach

The goal is to rearrange a collection of numbers so that they alternate between being smaller and larger than their neighbors. The brute force method tries out every possible order of the numbers until it finds one that satisfies this alternating pattern.

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

Start by listing out all the possible arrangements of the given numbers.
For each of these arrangements, check if it follows the wiggle sort pattern: the first number is smaller than the second, the second is larger than the third, the third is smaller than the fourth, and so on.
If an arrangement does follow the wiggle sort pattern, keep it as a potential solution.
After checking all the possible arrangements, pick any of the ones that worked. If none worked, it means something went wrong in your setup.

Code Implementation

import itertools

def wiggle_sort_brute_force(numbers):
    all_permutations = list(itertools.permutations(numbers))

    valid_permutations = []
    for permutation in all_permutations:
        is_wiggle_sort = True
        # Check if the current permutation satisfies the wiggle sort property

        for index in range(len(permutation) - 1):
            if index % 2 == 0:
                if permutation[index] >= permutation[index + 1]:
                    is_wiggle_sort = False
                    break
            else:
                if permutation[index] <= permutation[index + 1]:
                    is_wiggle_sort = False
                    break

        if is_wiggle_sort:
            valid_permutations.append(list(permutation))

    # If there are valid permutations, return the first one.
    if valid_permutations:
        return valid_permutations[0]
    else:
        return None

Big(O) Analysis

Time Complexity

O(n! * n) – The algorithm generates all permutations of the input array of size n. Generating all permutations has a time complexity of O(n!). For each permutation generated, the algorithm checks if it satisfies the wiggle sort condition. Checking a single permutation requires iterating through the array, which takes O(n) time. Therefore, the overall time complexity is O(n! * n), as we generate n! permutations and check each in O(n) time.

Space Complexity

O(N!) – The brute force approach generates all possible permutations of the input numbers. To store these permutations, which are candidate solutions, the algorithm implicitly uses space to hold each permutation. In the worst-case scenario, the algorithm could generate N! permutations before finding a valid wiggle sort, where N is the number of elements in the input array. Therefore, the space complexity for storing these permutations is proportional to the number of permutations, which is O(N!).

Optimal Solution

Approach

The key to solving this problem efficiently is realizing we only care about the relative order, not the exact values. We can cleverly arrange numbers by focusing on their median, dividing the problem into sorting elements around it. This approach avoids unnecessary comparisons and leads to a faster solution.

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

First, find the middle value of the entire collection of numbers.
Imagine two separate groups: one for smaller numbers and one for larger numbers. Some numbers might equal the middle value, and we'll deal with them carefully.
Place the larger numbers in every other position, starting from the second position from the end and moving backwards. If you run out of larger numbers, that's okay.
Place the smaller numbers in the remaining positions, starting from the very beginning and moving forward. Again, it's okay if you run out.
By strategically placing the larger and smaller values in this order, we guarantee the wiggle pattern: smaller, larger, smaller, larger and so on.

Code Implementation

def wiggle_sort_two(numbers):
    numbers_length = len(numbers)
    median_value = sorted(numbers)[numbers_length // 2]

    # This is the core wiggle sort logic.
    lesser_index = 0
    greater_index = (numbers_length - 1) // 2 + 1

    temp_array = [0] * numbers_length

    for index in range(numbers_length):
        if numbers[index] > median_value:
            temp_array[greater_index] = numbers[index]
            greater_index += 2
            if greater_index >= numbers_length:
                greater_index = 1

    for index in range(numbers_length):
        if numbers[index] < median_value:
            temp_array[lesser_index] = numbers[index]
            lesser_index += 2

    for index in range(numbers_length):
        if numbers[index] == median_value:
            temp_array[lesser_index] = numbers[index]
            lesser_index += 2
            if lesser_index >= numbers_length:
                lesser_index = 1

    # Overwrite the original array
    for index in range(numbers_length):
        numbers[index] = temp_array[index]

def wiggle_sort_two_in_place(numbers):
    numbers_length = len(numbers)
    median_value = sorted(numbers)[numbers_length // 2]

    lesser_index = 0
    greater_index = (numbers_length - 1) // 2 + 1
    
    temp_array = [0] * numbers_length

    for index in range(numbers_length):
        if numbers[index] > median_value:
            temp_array[greater_index] = numbers[index]
            greater_index += 2
            if greater_index >= numbers_length:
                greater_index = 1
                
    for index in range(numbers_length):
        if numbers[index] < median_value:
            temp_array[lesser_index] = numbers[index]
            lesser_index += 2
            
    for index in range(numbers_length):
        if numbers[index] == median_value:
            temp_array[lesser_index] = numbers[index]
            lesser_index += 2
            if lesser_index >= numbers_length:
                lesser_index = 1
                
    for index in range(numbers_length):
        numbers[index] = temp_array[index]

def wiggle_sort(nums):
    nums.sort()
    half = len(nums[::2])
    nums[::2], nums[1::2] = nums[:half][::-1], nums[half:][::-1]

Big(O) Analysis

Time Complexity

O(n log n) – The described approach involves first finding the median. A typical efficient algorithm for finding the median of n numbers is sorting, which generally takes O(n log n) time. The subsequent steps of placing larger and smaller numbers into their respective positions each iterate through the array once, contributing O(n) time. Therefore, the dominant factor in the overall time complexity is the median finding step via sorting. The overall complexity is O(n log n) + O(n), which simplifies to O(n log n).

Space Complexity

O(1) – The provided plain English explanation of the Wiggle Sort II algorithm focuses on rearranging the input array in place, primarily through comparisons and swaps. Although it mentions concepts like grouping numbers into larger and smaller piles relative to the median, the explanation does not describe explicitly creating any auxiliary data structures like temporary arrays, lists, or hash maps to store these groups. Therefore, the algorithm operates with a constant amount of extra space, independent of the input size N.

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty array or raise an exception (IllegalArgumentException) based on the problem's constraints.
Array with one element	Return the array as is, since it already satisfies the wiggle property.
Array with two elements where nums[0] == nums[1]	Swap them to satisfy the wiggle condition nums[0] < nums[1] (or nums[0] > nums[1] depending on the even/odd position).
Array with all elements identical	The standard wiggle sort algorithm might get stuck in an infinite loop or produce incorrect results, requiring careful handling typically solved via median selection techniques.
Array with highly skewed distribution (e.g., many small values, few large values)	This can negatively affect the efficiency of some sorting-based approaches, potentially leading to suboptimal median selection, so median-of-medians might be helpful.
Large input array potentially causing integer overflow during calculations (if array elements are large)	Use long data type to avoid integer overflow during arithmetic operations or compare array elements directly if possible.
Array containing negative numbers and zeros	The solution should correctly handle both negative numbers and zeros within the comparison and swapping logic.
Multiple valid wiggle sort orderings exist	The algorithm should produce one valid wiggle sort ordering, even if multiple solutions are possible, because the question asks for a valid ordering.