# Check if the given array is same as its inverse permutation

Given an array arr[] consisting of integers in the range [1, N], the task is to determine whether the Inverse Permutation of the given array is same as the given array.

An inverse permutation is a permutation obtained by inserting the position of all elements at the position equal to the respective values of the element in the array.
Illustration:
arr[] = {2, 4, 1, 3, 5}
The inverse permutation of the array will be equal to {3, 1, 4, 2, 5}

Examples:

Input: N = 4, arr[] = {1, 4, 3, 2}
Output: Yes
Explanation:
The inverse permutation of the given array is {1, 4, 3, 2} which is same as the given array.
Input: N = 5, arr[] = {2, 3, 4, 5, 1}
Output: No
Explanation:
The inverse permutation of the given array is {5, 1, 2, 3, 4} which is not the same as the given array.

Method 1 :

In this method, we will generate the inverse permutation of the array, then check if it is same as the original array.

Follow the steps below to solve the problem:

• Find the inverse permutation of the given array.
• Check, if the generated array is same as the original array.
• If both are same, then print Yes. Otherwise, print No.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement ` `// the above approach ` `#include ` `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

 `// 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

 `# 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#

 `// 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`

## Javascript

 ``

Output

`Yes`

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

Method 2 :

In this method, we take elements one by one and check for elements if at (arr[i] -1) index i+1 is present or not . If not then the inverse permutation of the given array is not the same as the array.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement` `// the above approach` `#include ` `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;` `}`

## Java

 `/*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

## Python3

 `# 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`

## C#

 `// 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` `bool` `inverseEqual(``int``[] arr,``int` `n){`   `    ``// Check the if inverse permutation is not same` `    ``for``(``int` `i=0;i

## Javascript

 ``

Output

`Yes`

Time Complexity: O(N)

Auxiliary Space: O(1)

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