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Shuffle the Array

Easy

Asked by:

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Topics:

Arrays

Given the array nums consisting of 2n elements in the form [x₁,x₂,...,x_n,y₁,y₂,...,y_n].

Return the array in the form [x₁,y₁,x₂,y₂,...,x_n,y_n].

Example 1:

Input: nums = [2,5,1,3,4,7], n = 3
Output: [2,3,5,4,1,7] 
Explanation: Since x₁=2, x₂=5, x₃=1, y₁=3, y₂=4, y₃=7 then the answer is [2,3,5,4,1,7].

Example 2:

Input: nums = [1,2,3,4,4,3,2,1], n = 4
Output: [1,4,2,3,3,2,4,1]

Example 3:

Input: nums = [1,1,2,2], n = 2
Output: [1,2,1,2]

Constraints:

1 <= n <= 500
nums.length == 2n
1 <= nums[i] <= 10^3

Shuffle the Array

Easy

Asked by:

14 views

Topics:

Arrays

Given the array nums consisting of 2n elements in the form [x₁,x₂,...,x_n,y₁,y₂,...,y_n].

Return the array in the form [x₁,y₁,x₂,y₂,...,x_n,y_n].

Example 1:

Input: nums = [2,5,1,3,4,7], n = 3
Output: [2,3,5,4,1,7] 
Explanation: Since x₁=2, x₂=5, x₃=1, y₁=3, y₂=4, y₃=7 then the answer is [2,3,5,4,1,7].

Example 2:

Input: nums = [1,2,3,4,4,3,2,1], n = 4
Output: [1,4,2,3,3,2,4,1]

Example 3:

Input: nums = [1,1,2,2], n = 2
Output: [1,2,1,2]

Constraints:

1 <= n <= 500
nums.length == 2n
1 <= nums[i] <= 10^3

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 expected range of values for the integers within the input array?
Can I assume that the input array `nums` will always have an even length, and that `n` will always be exactly half the length of `nums`?
What should I return if the input array is null or empty?
Is the input array guaranteed to be valid, or should I handle cases where `n` is not a valid midpoint (e.g., `n` is negative or larger than half the array length)?
Is there a preference for modifying the input array in-place versus creating a new array to store the shuffled result?

Brute Force Solution

Approach

The goal is to mix up the items in the list in a random order. The most straightforward approach involves trying out all possible arrangements and picking one at random. This ensures that any possible ordering has a chance of happening.

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

First, make a list of every single possible arrangement of the items.
Then, simply pick one of these arrangements at random to be the final shuffled list.

Code Implementation

import random
import itertools

def shuffle_array_brute_force(input_array):
    # Generate all possible permutations of the input array
    all_permutations = list(itertools.permutations(input_array))

    # Check if the array is empty, if so return it
    if not input_array:
        return input_array

    # Select a random permutation from the list of all permutations
    random_index = random.randint(0, len(all_permutations) - 1)

    # Return the shuffled array.
    shuffled_array = list(all_permutations[random_index])
    return shuffled_array

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n!) – The approach involves generating all possible permutations of the array. The number of permutations for an array of size n is n! (n factorial). Since we must generate all these permutations, the time complexity is at least O(n!). Picking a random permutation from the generated list can be done in O(1), but the permutation generation dominates the complexity. Therefore, the overall time complexity is O(n!).

Space Complexity

O(N!) – The provided algorithm generates all possible permutations of the input array of size N. This involves creating a list of all possible arrangements. The number of permutations for an array of size N is N! (N factorial). Therefore, the auxiliary space required to store all these permutations is proportional to N!.

Optimal Solution

Approach

The goal is to mix the first half of a list with the second half in a specific pattern. Instead of creating entirely new lists, the trick is to rearrange the original list directly to save effort and space.

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

Imagine you have two groups of items, A and B, and you want to merge them alternately: A, B, A, B, and so on.
Starting with the first item in the list, take the first item from group A and place it where it should go in the shuffled list.
Then, take the first item from group B and place it where it should go.
Continue alternating between the two groups, A and B, picking items and placing them into their correct shuffled positions.
Since you're rearranging items in the original list, you have to be careful not to lose any items. When moving an item, temporarily store the item that was originally in its new spot.
By carefully placing the items in order and temporarily holding any item that gets overwritten, we can make sure all items are shuffled correctly in one sweep.

Code Implementation

def shuffle_the_array(nums, number_of_pairs):
    list_length = len(nums)

    # Iterate through the first half of the array.
    for i in range(number_of_pairs):
        current_index = i
        current_value = nums[current_index]

        # Keep swapping until the element is in the correct position.
        while current_index < list_length:

            # Calculate target index in shuffled array.
            if current_index < number_of_pairs:
                target_index = 2 * current_index
            else:
                target_index = 2 * (current_index - number_of_pairs) + 1

            # If the element is already in the correct position, break.
            if target_index == i and current_index >= number_of_pairs:
                break

            # Store the value at the target index temporarily.
            temp_value = nums[target_index]
            nums[target_index] = current_value

            # Move to the target index for the next swap.
            current_index = target_index
            current_value = temp_value

    return nums

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The described algorithm iterates through the array of n elements. For each element, it performs a constant number of operations: moving the element to its correct shuffled position and temporarily storing the overwritten element. Since the work done for each element is constant and we iterate through all n elements once, the time complexity is directly proportional to n, hence O(n).

Space Complexity

O(1) – The plain English explanation describes an in-place shuffling algorithm. While rearranging the list, the explanation mentions temporarily storing an item that was originally in its new spot. This implies using a single temporary variable to hold the value being swapped. No other data structures that scale with the input size N are used. Therefore, the algorithm uses constant auxiliary space.

Edge Cases

Case	How to Handle
nums is null	Throw an IllegalArgumentException or return null to indicate invalid input.
nums is empty	Return an empty array immediately as there is nothing to shuffle.
n is zero	Return the original array as no shuffling is needed (or throw exception).
nums.length is not equal to 2*n	Throw an IllegalArgumentException to indicate that the input does not meet the problem's constraints.
Large input size (n is very large)	Ensure the solution's space and time complexity are efficient enough to handle large arrays without exceeding memory or time limits.
nums contains negative numbers	The provided algorithm should work correctly with negative numbers, but verify the correctness of any arithmetic operations if present.
nums contains zero	The algorithm should handle zero values in the input array without any issues.
n is very large such that 2*n causes integer overflow	Check for potential integer overflow during calculations involving n, and use a larger data type if necessary or validate input range.

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 expected range of values for the integers within the input array?
Can I assume that the input array `nums` will always have an even length, and that `n` will always be exactly half the length of `nums`?
What should I return if the input array is null or empty?
Is the input array guaranteed to be valid, or should I handle cases where `n` is not a valid midpoint (e.g., `n` is negative or larger than half the array length)?
Is there a preference for modifying the input array in-place versus creating a new array to store the shuffled result?

Brute Force Solution

Approach

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

First, make a list of every single possible arrangement of the items.
Then, simply pick one of these arrangements at random to be the final shuffled list.

Code Implementation

import random
import itertools

def shuffle_array_brute_force(input_array):
    # Generate all possible permutations of the input array
    all_permutations = list(itertools.permutations(input_array))

    # Check if the array is empty, if so return it
    if not input_array:
        return input_array

    # Select a random permutation from the list of all permutations
    random_index = random.randint(0, len(all_permutations) - 1)

    # Return the shuffled array.
    shuffled_array = list(all_permutations[random_index])
    return shuffled_array

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:

Imagine you have two groups of items, A and B, and you want to merge them alternately: A, B, A, B, and so on.
Starting with the first item in the list, take the first item from group A and place it where it should go in the shuffled list.
Then, take the first item from group B and place it where it should go.
Continue alternating between the two groups, A and B, picking items and placing them into their correct shuffled positions.
Since you're rearranging items in the original list, you have to be careful not to lose any items. When moving an item, temporarily store the item that was originally in its new spot.
By carefully placing the items in order and temporarily holding any item that gets overwritten, we can make sure all items are shuffled correctly in one sweep.

Code Implementation

def shuffle_the_array(nums, number_of_pairs):
    list_length = len(nums)

    # Iterate through the first half of the array.
    for i in range(number_of_pairs):
        current_index = i
        current_value = nums[current_index]

        # Keep swapping until the element is in the correct position.
        while current_index < list_length:

            # Calculate target index in shuffled array.
            if current_index < number_of_pairs:
                target_index = 2 * current_index
            else:
                target_index = 2 * (current_index - number_of_pairs) + 1

            # If the element is already in the correct position, break.
            if target_index == i and current_index >= number_of_pairs:
                break

            # Store the value at the target index temporarily.
            temp_value = nums[target_index]
            nums[target_index] = current_value

            # Move to the target index for the next swap.
            current_index = target_index
            current_value = temp_value

    return nums

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
nums is null	Throw an IllegalArgumentException or return null to indicate invalid input.
nums is empty	Return an empty array immediately as there is nothing to shuffle.
n is zero	Return the original array as no shuffling is needed (or throw exception).
nums.length is not equal to 2*n	Throw an IllegalArgumentException to indicate that the input does not meet the problem's constraints.
Large input size (n is very large)	Ensure the solution's space and time complexity are efficient enough to handle large arrays without exceeding memory or time limits.
nums contains negative numbers	The provided algorithm should work correctly with negative numbers, but verify the correctness of any arithmetic operations if present.
nums contains zero	The algorithm should handle zero values in the input array without any issues.
n is very large such that 2*n causes integer overflow	Check for potential integer overflow during calculations involving n, and use a larger data type if necessary or validate input range.