How do you solve Shuffle the Array on LeetCode?

Shuffle the Array (LeetCode #1470) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: The brute-force approach involves creating a new array and manually inserting elements from the two halves of the original array. This is simple but inefficient as it requires multiple iterations.

What algorithm or data structure is used to solve Shuffle the Array?

Shuffle the Array uses the following concepts: Array. The recommended patterns to study are: .

Shuffle the Array — LeetCode #1470 (Easy)

Tags: Array

Related patterns:

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute-force approach involves creating a new array and manually inserting elements from the two halves of the original array. This is simple but inefficient as it requires multiple iterations.

The time complexity is O(n²) because of the repeated appending in a loop, which can lead to multiple iterations over the array.

Step 1: Initialize an empty array result of size 2n.
Step 2: Iterate through the first half of nums and the second half, inserting elements alternately into result.
Step 3: Return the result array.

Input: nums = [2,5,1,3,4,7], n = 3 1. Initialize result = [] 2. i = 0: result = [2] 3. i = 0: result = [2, 3] 4. i = 1: result = [2, 3, 5] 5. i = 1: result = [2, 3, 5, 4] 6. i = 2: result = [2, 3, 5, 4, 1] 7. i = 2: result = [2, 3, 5, 4, 1, 7] Final output: [2, 3, 5, 4, 1, 7]

#1470

Shuffle the Array

Easy

Array↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

The brute-force approach involves creating a new array and manually inserting elements from the two halves of the original array. This is simple but inefficient as it requires multiple iterations.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty array result of size 2n.
2Step 2: Iterate through the first half of nums and the second half, inserting elements alternately into result.
3Step 3: Return the result array.

solution.py6 lines

1def shuffle(nums, n):
2    result = []
3    for i in range(n):
4        result.append(nums[i])
5        result.append(nums[i + n])
6    return result

ℹ

Complexity note: The time complexity is O(n²) because of the repeated appending in a loop, which can lead to multiple iterations over the array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.