Constructing Palindromic Arrays with First Element as X

Given two integers N and X (1 <= X <= N). The task is to find a permutation P of [1,2,3…N] with the first element as X, such that the difference array Diff[], where Diff_i = P_(i+1) – P_i for all i (1 <= i < N), forms a palindrome. If no such permutation exists, output -1.

Examples:

Input: N = 4, X = 3
Output: P[] = {3, 1, 4, 2}
Explanation: If P[] = {3, 1, 4, 2}, Then:

• First element of P[] is equal to X = 3, which is true.
• Calculating Diff[] for all i (1 <= i < N) gives = {(P₂-P₁), (P₃-P₂), (P₄-P₃)} = {(1-3), (4-1), (2-4)} = {-2, 3, -2}, which is palindrome.

Thus, P[] follows all the conditions, having length N and all elements are from the range [1, N].

Input: N = 3, X = 2
Output: -1
Explanation: It can be verified that no such P[] exists for the given values of N and X.

Approach:

The problem is observation based. The main idea is:

• If N is odd and X is equal to (N + 1)/2, then no such P[] exists.
• Otherwise, a possible permutation P[] will always be there. Which can be obtained by following below sequence:
  • First, we have to print the X.
  • Then print all numbers from 1 to N except X and (N – X + 1).
  • Finally print (N – X + 1).

Step-by-step approach:

• Check Special Case:
  • If N is odd and X is equal to (N + 1) / 2, output -1.
• Generate Permutation:
  • Calculate Y = N – X + 1.
  • Print X.
  • Print all numbers from 1 to N except X and Y.
  • Print Y.

Below is the implementation of the above approach:

C++

// C++ code to implement the approach
#include <iostream>
using namespace std;

// Method to output P[], If it is possible
void Find_permutation(int N, int X)
{
    // First condition, where N is odd and X is equal to
    // (N+1)/2
    if (N % 2 == 1 && X == (N + 1) / 2) {
        cout << -1;
        return;
    }

    // Else we are making the following arrangement by code
    // below: First, we have to print the X. Then print all
    // numbers from 1 to N except X and (N - X + 1). Finally
    // print (N - X + 1).
    int Y = N - X + 1;

    // First element of arrangement
    cout << X << " ";

    // printing all numbers from 1 to N except X and (N - X
    // + 1).
    for (int i = 1; i <= N; ++i) {
        if (i == X || i == Y)
            continue;
        cout << i << " ";
    }

    // Finally printing (N - X + 1)
    cout << Y << " ";
}

// Driver Function
int main()
{
    // Input
    int N = 4, X = 4;

    // Function_call
    Find_permutation(N, X);
    return 0;
}

// This code is contributed by ragul21

Java

// Java code to implement the approach
import java.util.*;

// Driver Class
public class GFG {
    // Driver Function
    public static void main(String[] args)
    {
        // Input
        int N = 4, X = 4;

        // Function_call
        Find_permutation(N, X);
    }

    // Method to output P[], If it is possible
    public static void Find_permutation(int N, int X)
    {

        //First condition, where N is odd and X is equal to
        // (N+1)/2
        if (N % 2 == 1 && X == (N + 1) / 2) {
            System.out.println(-1);
            return;
        }

        //Else we are making the following arrangement by code below:
        //First, we have to print the X.
        //Then print all numbers from 1 to N except X and (N - X + 1).
        //Finally print (N - X + 1).
        int Y = N - X + 1;

        //First element of arrangement
        System.out.print(X + " ");

        // printing all numbers from 1 to N except X and (N - X + 1).
        for (int i = 1; i <= N; ++i) {
            if (i == X || i == Y)
                continue;
            System.out.print(i + " ");
        }

        //Finally printing (N - X + 1)
        System.out.print(Y + " ");
    }
}

Python3

# Method to output P[], If it is possible
def find_permutation(N, X):
    # First condition, where N is odd and X is equal to
    # (N+1)/2
    if N % 2 == 1 and X == (N + 1) // 2:
        print(-1)
        return

    # Else we are making the following arrangement by code
    # below: First, we have to print the X. Then print all
    # numbers from 1 to N except X and (N - X + 1). Finally
    # print (N - X + 1).
    Y = N - X + 1

    # First element of arrangement
    print(X, end=" ")

    # printing all numbers from 1 to N except X and (N - X
    # + 1).
    for i in range(1, N + 1):
        if i == X or i == Y:
            continue
        print(i, end=" ")

    # Finally printing (N - X + 1)
    print(Y, end=" ")

# Driver Function
if __name__ == "__main__":
    # Input
    N = 4
    X = 4

    # Function call
    find_permutation(N, X)

# This code is contributed by akshitaguprzj3

C#

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

class GFG
{
    // Method to output P[], If it is possible
    static void FindPermutation(int N, int X)
    {
        // First condition, where N is odd and X is equal to
        // (N+1)/2
        if (N % 2 == 1 && X == (N + 1) / 2)
        {
            Console.WriteLine(-1);
            return;
        }

        // Else we are making the following arrangement by code
        // below: First, we have to print the X. Then print all
        // numbers from 1 to N except X and (N - X + 1). Finally
        // print (N - X + 1).
        int Y = N - X + 1;

        // First element of arrangement
        Console.Write(X + " ");

        // printing all numbers from 1 to N except X and (N - X
        // + 1).
        for (int i = 1; i <= N; ++i)
        {
            if (i == X || i == Y)
                continue;
            Console.Write(i + " ");
        }

        // Finally printing (N - X + 1)
        Console.Write(Y + " ");
    }

    // Driver Function
    static void Main()
    {
        // Input
        int N = 4, X = 4;

        // Function_call
        FindPermutation(N, X);
    }
}

Javascript

// Method to output P[], If it is possible
function findPermutation(N, X) {
    // First condition, where N is odd and X is equal to (N+1)/2
    if (N % 2 === 1 && X === Math.floor((N + 1) / 2)) {
        console.log(-1);
        return;
    }

    // Else we are making the following arrangement by code below:
    // First, we have to print the X.
    // Then print all numbers from 1 to N except X and (N - X + 1).
    // Finally print (N - X + 1).
    const Y = N - X + 1;

    // First element of arrangement
    process.stdout.write(X + " ");

    // Printing all numbers from 1 to N except X and (N - X + 1).
    for (let i = 1; i <= N; ++i) {
        if (i === X || i === Y)
            continue;
        process.stdout.write(i + " ");
    }

    // Finally printing (N - X + 1)
    console.log(Y + " ");
}

// Driver Function
function main() {
    // Input
    const N = 4, X = 4;

    // Function call
    findPermutation(N, X);
}


main();

// This code is contributed by shivamgupta310570

Output

4 2 3 1 

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