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Create Array of distinct components the place odd listed components 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 shouldn’t be a number of of Ai-1 when i is even
  • All Ai are pairwise distinct
  • 1 ≤ Ai ≤ 2⋅N

Be aware: 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 legitimate array as a result of:
A1 = 3 is a number of of A2 = 6
A2 = 6 shouldn’t 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

Method: The issue might be solved based mostly on the next statement:

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 straightforward 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 components of the above sequence are distinctive.
  • Ai is a number of of Ai-1 might be trivially verified to be true for all odd
    Ai shouldn’t be a number of of Ai-1 holds true for all even i.

Thus, the offered sequence fulfills all necessities of the issue, and is due to this fact legitimate!

Comply with the beneath steps to unravel the issue:

  • Initialize a variable oddElement = (N / 2) + 1 for odd listed components.
  • Initialize a variable evenElement = oddElement * 2 for even listed components.
  • 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.
       

Beneath is the implementation of the above strategy.

Java

  

import java.io.*;

import java.util.*;

  

public class GFG {

  

    

    public static void discover(int N)

    {

        int oddElement = N - (int)Math.ground(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 fundamental(String[] args)

    {

        int N = 4;

  

        

        discover(N);

    }

}

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

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