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Java Beginner Level Arrays Mock Interview

Question 2 of 13 for our Java Beginner Level Arrays Mock Interview

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Question 2 of 13

Create a solution to return a rearranged array.

This interview question concentrates on Java functions.

You are given array nums. It contains 2*n elements in form of-
[x1, x2, x3, …, xn, y1, y2, y3, …, yn]
Rearrange the given array to make in form of-
[x1, y1, x2, y2, x3, y3, …, xn, yn]
You have to return the solution in a new array.

/*Example*/

nums = [1,2,3,4,4,3,2,1], n = 4
expected = [1,4,2,3,3,2,4,1]

Solution:

We need to place the first n number to even indices (0-indexed) of the result array, and place next n elements to odd indices. This makes the solution pretty straightforward.

Create a new result array of size 2*n. Copy first n number to even indices, that is 2 * i and other n elements to odd indices, that is 2 * i + 1.

class Solution {
    public int[] shuffle(int[] nums, int n) {
        int result[] = new int[2 * n];
    
        for (int i = 0; i < n; i++) {
            // first n elements to even indices
            result[2 * i] = nums[i];
 
            // other n elements to even indices
            result[2 * i + 1] = nums[n + i];
        }
    
        return result;
    }

}

Time complexity- O(n)
Space complexity- O(1)

Written by on June 27th, 2021

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How to Answer: Create a solution to return a rearranged array.

Advice and answer examples written specifically for a Java Beginner Level Arrays job interview.

  • 2. Create a solution to return a rearranged array.

      This interview question concentrates on Java functions.

      You are given array nums. It contains 2*n elements in form of-
      [x1, x2, x3, …, xn, y1, y2, y3, …, yn]
      Rearrange the given array to make in form of-
      [x1, y1, x2, y2, x3, y3, …, xn, yn]
      You have to return the solution in a new array.

      /*Example*/
      
      nums = [1,2,3,4,4,3,2,1], n = 4
      expected = [1,4,2,3,3,2,4,1]

      Solution:

      We need to place the first n number to even indices (0-indexed) of the result array, and place next n elements to odd indices. This makes the solution pretty straightforward.

      Create a new result array of size 2*n. Copy first n number to even indices, that is 2 * i and other n elements to odd indices, that is 2 * i + 1.

      class Solution {
          public int[] shuffle(int[] nums, int n) {
              int result[] = new int[2 * n];
          
              for (int i = 0; i < n; i++) {
                  // first n elements to even indices
                  result[2 * i] = nums[i];
       
                  // other n elements to even indices
                  result[2 * i + 1] = nums[n + i];
              }
          
              return result;
          }
      
      }

      Time complexity- O(n)
      Space complexity- O(1)

      Written by S. Kumar on June 27th, 2021