Compruebe si la array dada es igual a su permutación inversa

Dada una array arr[] que consta de números enteros en el rango [1, N] , la tarea es determinar si la permutación inversa de la array dada es la misma que la array dada. 

Una permutación inversa es una permutación obtenida al insertar la posición de todos los elementos en la posición igual a los valores respectivos del elemento en la array.
Ilustración: 
arr[] = {2, 4, 1, 3, 5} 
La permutación inversa de la array será igual a {3, 1, 4, 2, 5} 

Ejemplos: 

Entrada: N = 4, arr[] = {1, 4, 3, 2} 
Salida: Sí 
Explicación: 
La permutación inversa de la array dada es {1, 4, 3, 2} que es igual a la array dada.
Entrada: N = 5, arr[] = {2, 3, 4, 5, 1} 
Salida: No 
Explicación: 
La permutación inversa de la array dada es {5, 1, 2, 3, 4} que no es lo mismo como la array dada. 

Método 1 :

 En este método, generaremos la permutación inversa de la array, luego verificaremos si es igual a la array original.

Siga los pasos a continuación para resolver el problema: 

• Encuentre la permutación inversa de la array dada.
• Verifique si la array generada es la misma que la array original.
• Si ambos son iguales, imprima Sí . De lo contrario , imprima No.

A continuación se muestra la implementación del enfoque anterior: 

// C++ Program to implement
// the above approach
#include <iostream>
using namespace std;
 
// Function to check if the inverse
// permutation of the given array is
// same as the original array
void inverseEqual(int arr[], int n)
{
 
    // Stores the inverse
    // permutation
    int brr[n];
 
    // Generate the inverse permutation
    for (int i = 0; i < n; i++) {
        int present_index = arr[i] - 1;
        brr[present_index] = i + 1;
    }
 
    // Check if the inverse permutation
    // is same as the given array
    for (int i = 0; i < n; i++) {
        if (arr[i] != brr[i]) {
            cout << "No" << endl;
            return;
        }
    }
 
    cout << "Yes" << endl;
}
 
// Driver Code
int main()
{
 
    int n = 4;
    int arr[n] = { 1, 4, 3, 2 };
 
    inverseEqual(arr, n);
 
    return 0;
}

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to check if the inverse
// permutation of the given array is
// same as the original array
static void inverseEqual(int arr[], int n)
{
     
    // Stores the inverse
    // permutation
    int[] brr = new int[n];
 
    // Generate the inverse permutation
    for(int i = 0; i < n; i++)
    {
        int present_index = arr[i] - 1;
        brr[present_index] = i + 1;
    }
 
    // Check if the inverse permutation
    // is same as the given array
    for(int i = 0; i < n; i++)
    {
        if (arr[i] != brr[i])
        {
            System.out.println("No");
            return;
        }
    }
    System.out.println("Yes");
}
 
// Driver code
public static void main(String[] args)
{
    int n = 4;
    int[] arr = { 1, 4, 3, 2 };
 
    inverseEqual(arr, n);
}
}
 
// This code is contributed by offbeat

# Python3 program to implement
# the above approach
 
# Function to check if the inverse
# permutation of the given array is
# same as the original array
def inverseEqual(arr, n):
     
    # Stores the inverse
    # permutation
    brr = [0] * n
     
    # Generate the inverse permutation
    for i in range(n):
        present_index = arr[i] - 1
        brr[present_index] = i + 1
         
    # Check if the inverse permutation
    # is same as the given array
    for i in range(n):
        if arr[i] != brr[i]:
            print("NO")
            return
             
    print("YES")
     
# Driver code
n = 4
arr = [ 1, 4, 3, 2 ]
 
inverseEqual(arr, n)
 
# This code is contributed by Stuti Pathak

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to check if the inverse
// permutation of the given array is
// same as the original array
static void inverseEqual(int []arr, int n)
{
     
    // Stores the inverse
    // permutation
    int[] brr = new int[n];
 
    // Generate the inverse permutation
    for(int i = 0; i < n; i++)
    {
        int present_index = arr[i] - 1;
        brr[present_index] = i + 1;
    }
 
    // Check if the inverse permutation
    // is same as the given array
    for(int i = 0; i < n; i++)
    {
        if (arr[i] != brr[i])
        {
            Console.WriteLine("No");
            return;
        }
    }
    Console.WriteLine("Yes");
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 4;
    int[] arr = { 1, 4, 3, 2 };
 
    inverseEqual(arr, n);
}
}
 
// This code is contributed by sapnasingh4991

<script>
 
// Javascript Program to implement
// the above approach
 
// Function to check if the inverse
// permutation of the given array is
// same as the original array
function inverseEqual(arr, n)
{
 
    // Stores the inverse
    // permutation
    var brr = Array(n).fill(0);
 
    // Generate the inverse permutation
    for (var i = 0; i < n; i++) {
        var present_index = arr[i] - 1;
        brr[present_index] = i + 1;
    }
 
    // Check if the inverse permutation
    // is same as the given array
    for (var i = 0; i < n; i++) {
        if (arr[i] != brr[i]) {
            document.write( "No" );
            return;
        }
    }
 
    document.write( "Yes" );
}
 
// Driver Code
var n = 4;
var arr = [ 1, 4, 3, 2 ];
inverseEqual(arr, n);
 
// This code is contributed by noob2000.
</script>

Yes

Complejidad temporal: O(N)  
Espacio auxiliar: O(N)

 Método 2:

En este método, tomamos los elementos uno por uno y verificamos si en (arr[i] -1) el índice i+1 está presente o no . De lo contrario, la permutación inversa de la array dada no es la misma que la array.

A continuación se muestra la implementación del enfoque anterior:

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if the inverse
// permutation of the given array is
// same as the original array
bool inverseEqual(int arr[], int n)
{
 
    // Check the if inverse permutation is not same
    for (int i = 0; i < n; i++)
        if (arr[arr[i] - 1] != i + 1)
            return false;
 
    return true;
}
 
// Driver Code
int main()
{
    int n = 4;
    int arr[n] = { 1, 4, 3, 2 };
 
    // Function Call
    cout << (inverseEqual(arr, n) ? "Yes" : "No");
 
    return 0;
}

/*package whatever //do not write package name here */
import java.io.*;
 
class GFG
{
 
  // Java program to implement
  // the above approach
 
  // Function to check if the inverse
  // permutation of the given array is
  // same as the original array
  static boolean inverseEqual(int[] arr,int n){
 
    // Check the if inverse permutation
    // is not same
    for(int i=0;i<n;i++){
      if (arr[arr[i] - 1] != i + 1)
        return false;
    }
 
    return true;
  }
 
  // Driver code
  public static void main(String args[])
  {
    int n = 4;
    int[] arr = { 1, 4, 3, 2 };
 
    // Function Call
    System.out.println((inverseEqual(arr, n)==true)?"Yes" : "No");
  }
}
 
// This Code is contributed by shinjanpatra

# Python program to implement
# the above approach
 
# Function to check if the inverse
# permutation of the given array is
# same as the original array
def inverseEqual(arr, n):
 
    # Check the if inverse permutation
    # is not same
    for i in range(n):
        if (arr[arr[i] - 1] != i + 1):
            return False
 
    return True
 
# Driver Code
n = 4
arr = [ 1, 4, 3, 2 ]
 
# Function Call
print("Yes" if (inverseEqual(arr, n)==True) else "No")
 
# This code is contributed by shinjanpatra

<script>
 
// Javascript program to implement
// the above approach
 
// Function to check if the inverse
// permutation of the given array is
// same as the original array
function inverseEqual(arr, n)
{
  
    // Check the if inverse permutation
    // is not same
    for (let i = 0; i < n; i++)
        if (arr[arr[i] - 1] != i + 1)
            return false;
  
    return true;
}
 
    // Driver Code
     
    let n = 4;
    let arr = [ 1, 4, 3, 2 ];
  
    // Function Call
    document.write(inverseEqual(arr, n) ? "Yes" : "No");
 
</script>

Yes

Complejidad de tiempo: O(N) 

Espacio Auxiliar: O(1)