1470. Shuffle the Array LeetCode Solution

In this guide, you will get 1470. Shuffle the Array LeetCode Solution with the best time and space complexity. The solution to Shuffle the Array problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Shuffle the Array solution in C++
Shuffle the Array solution in Java
Shuffle the Array solution in Python
Additional Resources

1470. Shuffle the Array LeetCode Solution image

Problem Statement of Shuffle the Array

Given the array nums consisting of 2n elements in the form [x1,x2,…,xn,y1,y2,…,yn].
Return the array in the form [x1,y1,x2,y2,…,xn,yn].

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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

1470. Shuffle the Array LeetCode Solution in C++

class Solution {
 public:
  vector<int> shuffle(vector<int>& nums, int n) {
    vector<int> ans;
    for (int i = 0; i < n; ++i) {
      ans.push_back(nums[i]);
      ans.push_back(nums[i + n]);
    }
    return ans;
  }
};
/* code provided by PROGIEZ */

1470. Shuffle the Array LeetCode Solution in Java

class Solution {
  public int[] shuffle(int[] nums, int n) {
    int[] ans = new int[2 * n];
    for (int i = 0; i < n; ++i) {
      ans[i * 2] = nums[i];
      ans[i * 2 + 1] = nums[i + n];
    }
    return ans;
  }
}
// code provided by PROGIEZ

1470. Shuffle the Array LeetCode Solution in Python

class Solution:
  def shuffle(self, nums: list[int], n: int) -> list[int]:
    ans = []
    for a, b in zip(nums[:n], nums[n:]):
      ans.append(a)
      ans.append(b)
    return ans
 # code by PROGIEZ