Problem Description

You are given an array nums that contains 2n elements arranged in a specific format: [x₁, x₂, ..., xₙ, y₁, y₂, ..., yₙ].

The array is divided into two halves:

• The first half contains elements x₁ through xₙ
• The second half contains elements y₁ through yₙ

Your task is to rearrange the array by interleaving elements from both halves. The resulting array should follow the pattern: [x₁, y₁, x₂, y₂, ..., xₙ, yₙ].

In other words, you need to take the first element from the first half, then the first element from the second half, then the second element from the first half, then the second element from the second half, and so on.

For example, if nums = [2, 5, 1, 3, 4, 7] and n = 3:

• First half: [2, 5, 1] (elements x₁, x₂, x₃)
• Second half: [3, 4, 7] (elements y₁, y₂, y₃)
• Result: [2, 3, 5, 4, 1, 7] (following the pattern x₁, y₁, x₂, y₂, x₃, y₃)

The solution uses Python's zip function to pair up elements from the first half (nums[:n]) with elements from the second half (nums[n:]), then flattens these pairs into a single list using a list comprehension.

Intuition

When we look at the input and output patterns, we can observe that we're essentially pairing elements from two halves of the array. The first element of the result comes from position 0, the second from position n, the third from position 1, the fourth from position n+1, and so on.

This pattern suggests we're alternating between picking elements from the first half and the second half of the original array. We can think of this as having two separate lists:

• First list: [x₁, x₂, ..., xₙ] (indices 0 to n-1)
• Second list: [y₁, y₂, ..., yₙ] (indices n to 2n-1)

Now we need to interleave these two lists. The natural way to do this is to create pairs: (x₁, y₁), (x₂, y₂), ..., (xₙ, yₙ), and then flatten these pairs into a single sequence.

In Python, the zip function is perfect for creating these pairs - it takes two sequences and pairs up elements at the same positions. So zip(nums[:n], nums[n:]) gives us the pairs we need. Then we just need to flatten these pairs, which can be elegantly done with a nested list comprehension: for each pair, we extract both elements in order.

The beauty of this approach is that it directly models our understanding of the problem - we're not manually tracking indices or doing complex arithmetic. We're simply saying "pair up corresponding elements from the two halves, then unpack those pairs in order."

Solution Approach

Following the simulation approach, we need to traverse through the first n elements and pair each element with its corresponding element from the second half of the array.

The implementation breaks down into these steps:

1. Split the array conceptually: We divide nums into two halves:

   • First half: nums[:n] contains elements from index 0 to n-1
   • Second half: nums[n:] contains elements from index n to 2n-1

2. Create pairs using zip: The zip(nums[:n], nums[n:]) function pairs up elements at the same positions from both halves:

   • It creates tuples like (nums[0], nums[n]), (nums[1], nums[n+1]), ..., (nums[n-1], nums[2n-1])
   • This gives us pairs (x₁, y₁), (x₂, y₂), ..., (xₙ, yₙ)

3. Flatten the pairs: We use a nested list comprehension to unpack each pair:

   [x for pair in zip(nums[:n], nums[n:]) for x in pair]

   • The outer loop for pair in zip(...) iterates through each tuple
   • The inner loop for x in pair extracts both elements from each tuple in order
   • This produces the sequence: x₁, y₁, x₂, y₂, ..., xₙ, yₙ

The time complexity is O(n) since we iterate through all 2n elements exactly once. The space complexity is O(n) for creating the result array (not counting the input array).

This solution elegantly uses Python's built-in functions to achieve the interleaving pattern without explicit index manipulation or complex loops.

Example Walkthrough

Let's walk through a small example with nums = [1, 3, 5, 2, 4, 6] where n = 3.

Step 1: Identify the two halves

• First half: nums[:3] = [1, 3, 5] (these are x₁, x₂, x₃)
• Second half: nums[3:] = [2, 4, 6] (these are y₁, y₂, y₃)

Step 2: Create pairs using zip When we call zip([1, 3, 5], [2, 4, 6]), it pairs elements at the same positions:

• Pair 1: (1, 2) - first element from each half
• Pair 2: (3, 4) - second element from each half
• Pair 3: (5, 6) - third element from each half

Step 3: Flatten the pairs Now we unpack each pair in sequence using the list comprehension:

[x for pair in zip([1, 3, 5], [2, 4, 6]) for x in pair]

Processing each pair:

• From pair (1, 2): we get 1, then 2
• From pair (3, 4): we get 3, then 4
• From pair (5, 6): we get 5, then 6

Final Result: [1, 2, 3, 4, 5, 6]

This achieves the desired interleaving pattern where we alternate between elements from the first half and second half: x₁, y₁, x₂, y₂, x₃, y₃.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the parameter passed to the function (which is half the length of the nums array).

• The slicing operations nums[:n] and nums[n:] each take O(n) time
• The zip() function iterates through both slices simultaneously in O(n) time
• The list comprehension iterates through all n pairs, and for each pair, it yields 2 elements, resulting in O(2n) = O(n) iterations
• Overall time complexity: O(n)

The space complexity is O(n), where n is the parameter passed to the function.

• The slices nums[:n] and nums[n:] create new lists, each requiring O(n) space
• The zip() object itself uses O(1) space as it's a generator
• The final list comprehension creates a new list of size 2n, which is O(n) space
• Overall space complexity: O(n) for the intermediate slices plus O(n) for the output list, which is O(n)

Note: In the reference answer, when it states "n is the length of the array nums", this appears to be the total length. In the code, the parameter n represents half of that length, but the complexity analysis remains O(n) relative to the parameter n.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Index Calculation in Manual Implementation

A common mistake when implementing this problem manually (without using zip) is miscalculating the indices when trying to pick elements from the second half of the array.

Incorrect approach:

def shuffle(self, nums: List[int], n: int) -> List[int]:
    result = []
    for i in range(n):
        result.append(nums[i])
        result.append(nums[i + n + 1])  # Wrong! Off-by-one error
    return result

Why it's wrong: The second half starts at index n, not n + 1. For element at position i in the first half, its corresponding element in the second half is at position i + n, not i + n + 1.

Correct manual implementation:

def shuffle(self, nums: List[int], n: int) -> List[int]:
    result = []
    for i in range(n):
        result.append(nums[i])
        result.append(nums[i + n])  # Correct index
    return result

2. Attempting In-Place Modification Without Proper Algorithm

Some might try to modify the array in-place by swapping elements, but this can lead to overwriting values before they're moved to their correct positions.

Problematic approach:

def shuffle(self, nums: List[int], n: int) -> List[int]:
    for i in range(n):
        # This overwrites values that haven't been processed yet
        nums[2 * i] = nums[i]
        nums[2 * i + 1] = nums[i + n]
    return nums

Why it fails: When you set nums[2 * i], you might be overwriting a value that hasn't been moved yet. For example, when i = 1, you're setting nums[2] = nums[1], but nums[2] hasn't been processed and moved to its correct position yet.

Solution: If in-place modification is required, you need a more sophisticated algorithm that tracks which elements have been moved, or use additional space to store intermediate values.

3. Misunderstanding the Problem Requirements

Some might interpret the problem as simply alternating between picking from the beginning and end of the array, rather than from two consecutive halves.

Wrong interpretation:

def shuffle(self, nums: List[int], n: int) -> List[int]:
    result = []
    for i in range(n):
        result.append(nums[i])
        result.append(nums[2*n - 1 - i])  # Taking from the end, not the second half
    return result

Correct understanding: The array is split into two consecutive halves [x₁...xₙ] and [y₁...yₙ], not alternating between start and end positions.