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Create Array of distinct parts the place odd listed parts are a number of of left neighbour

Create Array of distinct parts the place odd listed parts are a number of of left neighbour
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Given an integer N, the duty is to generate an array A[] of size N such that it satisfies the next situations for all 1 ≤ i ≤ N−1:

  • Ai is a number of of Ai-1 when i is odd
  • Ai will not be a number of of Ai-1 when i is even
  • All Ai are pairwise distinct
  • 1 ≤ Ai ≤ 2⋅N

Word: If there are a number of solutions print any of them.

    Examples:

Enter: N = 4
Output: 3 6 4 8 

Rationalization:  [3, 6, 4, 8] is a sound array as a result of:
A1 = 3 is a number of of A2 = 6
A2 = 6 will not be a number of of A3 = 4
A3 = 4 is a number of of A4 = 8.

Enter: N = 6
Output: 4 8 5 10 6 12

Strategy: The issue may be solved primarily based on the next remark:

Observations:

Let x = N − ⌈N / 2⌉ + 1. Then, the next sequence is legitimate: [x, 2⋅x, x + 1, 2⋅(x + 1), x + 2, …]

  • It’s simple to see that the weather at odd indices are in growing order from x→N. Equally, the weather at even indices are in growing order from 2⋅x→2⋅N (from 2⋅x→2⋅(N − 1) when N is odd).
  • Then, 2⋅x = 2⋅(N − ⌈N / 2⌉ + 1) > N implies the units {x, x + 1, …, N} and {2⋅x, 2⋅(x + 1), …, 2⋅N} are disjoint. Subsequently, all parts of the above sequence are distinctive.
  • Ai is a number of of Ai-1 may be trivially verified to be true for all odd
    Ai will not be a number of of Ai-1 holds true for all even i.

Thus, the supplied sequence fulfills all necessities of the issue, and is subsequently legitimate!

Observe the beneath steps to resolve the issue:

  • Initialize a variable oddElement = (N / 2) + 1 for odd listed parts.
  • Initialize a variable evenElement = oddElement * 2 for even listed parts.
  • Traverse a loop from 1 until N on i:
    • If i is odd print oddElement.
      • Assign evenElement = oddElement * 2.
      • Increment the evenElement.   
    • Else print the evenElement.
       

Under is the implementation of the above method.

Java

  

import java.io.*;

import java.util.*;

  

public class GFG {

  

    

    public static void discover(int N)

    {

        int oddElement = N - (int)Math.flooring(N / 2) + 1;

        int evenElement = 2 * oddElement;

  

        for (int i = 1; i <= N; i++) {

  

            

            if ((i % 2) != 0) {

                System.out.print(oddElement + " ");

                evenElement = 2 * oddElement;

                oddElement++;

            }

  

            

            else {

                System.out.print(evenElement + " ");

            }

        }

    }

  

    

    public static void most important(String[] args)

    {

        int N = 4;

  

        

        discover(N);

    }

}

Time Complexity: O(N)
Auxiliary House: O(1)

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