#1470

Easy

auto_awesomeArray-driven solution strategy

LeetCode Problem Workspace

Shuffle the Array

Shuffle the Array requires an efficient approach to rearrange elements using an array-driven solution strategy with two pointers.

array

Problem Statement

Given an array nums consisting of 2n elements, where the first half of the array represents x1, x2, ..., xn and the second half represents y1, y2, ..., yn, you are tasked with rearranging the array into the form [x1, y1, x2, y2, ..., xn, yn].

For example, given nums = [2,5,1,3,4,7] and n = 3, the output should be [2,3,5,4,1,7]. Your solution should be efficient and employ an array-driven approach to rearrange the elements in a single pass.

Examples

Example 1

Input: nums = [2,5,1,3,4,7], n = 3

Output: [2,3,5,4,1,7]

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 details omitted.

Example 3

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

Output: [1,2,1,2]

Example details omitted.

Constraints

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

Solution Approach

Two Pointers

Start by using two pointers, one at the beginning and another at the midpoint. Iterate through the array, pairing each element from the first half with the corresponding element from the second half, and construct the rearranged array.

In-Place Rearrangement

To achieve optimal space complexity, consider modifying the array in place. This avoids the need for additional space while still achieving the desired rearranged structure.

Efficient Handling of Multiple Pairs

Ensure that the two pointers efficiently handle all pairs of elements by incrementing both pointers in each iteration, ensuring that all elements are processed in the correct order.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution is O(n), where n is the size of the half array (i.e., the number of pairs). The space complexity is O(1) if done in-place, or O(n) if a new array is created.

What Interviewers Usually Probe

Can the candidate come up with a solution that processes the array in a single pass?
Does the candidate understand how to implement a two-pointer strategy?
Can the candidate handle space constraints by modifying the array in place?

Common Pitfalls or Variants

Common pitfalls

Failing to handle the rearrangement correctly and ending up with an incorrect order.
Using unnecessary extra space when the problem can be solved in-place.
Not maintaining the correct pairing between elements from the two halves of the array.

Follow-up variants

What if the array contains more than two halves? Extend the problem to more parts.
What if the array elements are not strictly consecutive or are unsorted?
What if the input array is already partially shuffled? How would you handle this case?

FAQ

What is the time complexity of the Shuffle the Array problem?

The time complexity is O(n), where n is the size of the half array, as each element is processed once.

Can I solve the problem in-place?

Yes, the problem can be solved in-place with O(1) space complexity by modifying the array while rearranging elements.

What is the optimal solution for the Shuffle the Array problem?

The optimal solution uses a two-pointer approach to rearrange the elements in a single pass, ensuring O(n) time complexity.

Can this problem be extended to more than two halves?

Yes, the approach can be generalized to handle more than two halves by using additional pointers for each half.

What are the common mistakes when solving the Shuffle the Array problem?

Common mistakes include failing to properly pair elements from the two halves and using unnecessary space for the solution.

terminal

Solution

Solution 1: Simulation

We traverse the indices $i$ in the range $[0, n)$. Each time, we take $\textit{nums}[i]$ and $\textit{nums}[i+n]$ and place them sequentially into the answer array.

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class Solution:
    def shuffle(self, nums: List[int], n: int) -> List[int]:
        return [x for pair in zip(nums[:n], nums[n:]) for x in pair]

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