Permutation of first N elements with absolute adjacent difference in increasing order

Given a positive integer N, the task is to construct a permutation from 1 to N such that the absolute difference of elements is in strictly increasing order. 

Note: N cannot be 0 or 1.

Examples:

Input: N = 10
Output: 6 5 7 4 8 3 9 2 10 1
Explanation: abs(6 - 5) i.e., 1 < abs(5 - 7) i.e., 2 < abs(7 - 4) i.e., 3 .... < abs(2 - 10) i.e., 8 < abs(10 - 1) i.e., 9

Input: 3
Output: 2 3 1 
Explanation: abs(2 - 3) = 1 and abs(3 - 1) = 2, 1 < 2 hence it is in strictly increasing order.

Approach: The problem can be solved based on the following observation:

Observation:

Let's say, you have the i =1 and j = N, the largest absolute difference made is by subtracting 1 and N = (N - 1)

Next Time, i increment by 1, i = 2 and j remains same i.e., N, So, the absolute difference is = (N - 2).
Next Time, i remains same i.e., 2 and j decrement by 1, j = N-1, So, the absolute difference is = (N - 1 - 2) = (N - 3).
Next Time, i increment by 1, i = 3 and j remains same i.e., N-1, So, the absolute difference is = (N - 1 - 3) = (N - 4).
Next Time, i remains same i.e., 3 and j decrement by 1, j = N-2, So, the absolute difference is = (N - 2 - 3) = (N - 5)......

Now, this way the series go, and at last two condition possible,

• When i = j + 1, [If N is odd], absolute difference = 1
• Or, j = i + 1, [If N is even], absolute difference = 1

So, this way the series become for given N, series = (N - 1), (N - 2), (N - 3), .... 3, 2, 1.

Follow the below steps to solve the problem:

• Initialize a pointer i = 1 and j = N.
• Declare an array of size N.
• Run a loop (using iterator x) from 0 to N - 1.
  • If x is even then set, arr[x] = i and increment i by 1.
  • Else then set, arr[x] = j and decrement j by 1.
• After executing the loop, print the array in reverse order.

Below is the implementation of the above approach:

// C++ code to implement the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to print valid permutation of N
void findPerm(int n)
{

    // Initialize the pointers
    int i = 1, j = n;

    // Declare an array of size N
    int arr[n];

    // Constructing the array
    for (int x = 0; x < n; x++) {
        if (x & 1)
            arr[x] = j--;
        else
            arr[x] = i++;
    }

    // Printing the array
    for (int x = (n - 1); x >= 0; x--) {
        cout << arr[x] << " ";
    }
}

// Driver Code
int main()
{
    int N = 10;

    // Function Call
    findPerm(N);
    return 0;
}

// Java code to implement the above approach

public class GFG {
    
// Function to print valid permutation of N
static void findPerm(int n)
{

    // Initialize the pointers
    int i = 1, j = n;

    // Declare an array of size N
    int arr[] = new int[n];

    // Constructing the array
    for (int x = 0; x < n; x++) {
        if ((x & 1) == 1)
            arr[x] = j--;
        else
            arr[x] = i++;
    }

    // Printing the array
    for (int x = (n - 1); x >= 0; x--) {
        System.out.print(arr[x]+" ");
    }
}

// Driver Code
public static void main (String[] args)
{
    int N = 10;

    // Function Call
    findPerm(N);
}

}

// This code is contributed by AnkThon

# python3 code to implement the above approach

# Function to print valid permutation of N
def findPerm(n):

    # Initialize the pointers
    i, j = 1, n

    # Declare an array of size N
    arr = [0 for _ in range(n)]

    # Constructing the array
    for x in range(0, n):
        if (x & 1):
            arr[x] = j
            j -= 1
        else:
            arr[x] = i
            i += 1

    # Printing the array
    for x in range(n-1, -1, -1):
        print(arr[x], end=" ")

# Driver Code
if __name__ == "__main__":

    N = 10

    # Function Call
    findPerm(N)

    # This code is contributed by rakeshsahni

// C# code to implement the above approach
using System;

public class GFG {

  // Function to print valid permutation of N
  static void findPerm(int n)
  {

    // Initialize the pointers
    int i = 1, j = n;

    // Declare an array of size N
    int[] arr = new int[n];

    // Constructing the array
    for (int x = 0; x < n; x++) {
      if ((x & 1) == 1)
        arr[x] = j--;
      else
        arr[x] = i++;
    }

    // Printing the array
    for (int x = (n - 1); x >= 0; x--) {
      Console.Write(arr[x] + " ");
    }
  }

  // Driver Code
  public static void Main(string[] args)
  {
    int N = 10;

    // Function Call
    findPerm(N);
  }
}

// This code is contributed by phasing17

<script>
 // Javascript code to implement the above approach
 
// Function to print valid permutation of N
function findPerm(n)
{

    // Initialize the pointers
    let i = 1, j = n;

    // Declare an array of size N
   let arr=new Array(n);

    // Constructing the array
    for (let x = 0; x < n; x++) {
        if (x & 1)
            arr[x] = j--;
        else
            arr[x] = i++;
    }

    // Printing the array
    for (let x = (n - 1); x >= 0; x--) {
        document.write(arr[x] + " ");
    }
}

// Driver Code
    let N = 10;

    // Function Call
    findPerm(N);
    
    // This code is contributed by satwik4409.
    </script>

6 5 7 4 8 3 9 2 10 1 

Time Complexity: O(N)
Auxiliary Space: O(N)