# Inverse Permutation

Given an array of size n of integers in range from 1 to n, we need to find the inverse permutation of that array.

An inverse permutation is a permutation which you will get by inserting position of an element at the position specified by the element value in the array. For better understanding, consider the following example:

Suppose we found element 4 at position 3 in an array, then in reverse permutation, we insert 3 (position of element 4 in the array) in position 4 (element value).

Basically, An inverse permutation is a permutation in which each number and the number of the place which it occupies is exchanged.

The array should contain element from 1 to array_size.

Example 1 :

```Input  = {1, 4, 3, 2}
Output = {1, 4, 3, 2}```

In this, For element 1 we insert position of 1 from arr1 i.e 1 at position 1 in arr2. For element 4 in arr1, we insert 2 from arr1 at position 4 in arr2. Similarly, for element 2 in arr1, we insert position of 2 i.e 4 in arr2.

```Example 2 :
Input  = {2, 3, 4, 5, 1}
Output = {5, 1, 2, 3, 4}```

In this example, for element 2 we insert position of 2 from arr1 in arr2 at position 2. similarly, we find the inverse permutation of other elements.
Consider an array arr having elements 1 to n.

Method 1: In this method, we take element one by one and check elements in increasing order and print the position of the element where we find that element.

Implementation:

## C++

 `// Naive CPP Program to find inverse permutation.` `#include ` `using` `namespace` `std;`   `// C++ function to find inverse permutations` `void` `inversePermutation(``int` `arr[], ``int` `size) {`   `  ``// Loop to select Elements one by one` `  ``for` `(``int` `i = 0; i < size; i++) {`   `    ``// Loop to print position of element` `    ``// where we find an element` `    ``for` `(``int` `j = 0; j < size; j++) {`   `      ``// checking the element in increasing order` `      ``if` `(arr[j] == i + 1) {`   `        ``// print position of element where` `        ``// element is in inverse permutation` `        ``cout << j + 1 << ``" "``;` `        ``break``;` `      ``}` `    ``}` `  ``}` `}`   `// Driver program to test above function` `int` `main() {` `  ``int` `arr[] = {2, 3, 4, 5, 1};` `  ``int` `size = ``sizeof``(arr) / ``sizeof``(arr[0]);` `  ``inversePermutation(arr, size);` `  ``return` `0;` `}`

## Java

 `// Naive java Program to find inverse permutation.` `import` `java.io.*;`   `class` `GFG {`   `    ``// java function to find inverse permutations` `    ``static` `void` `inversePermutation(``int` `arr[], ``int` `size)` `    ``{` `        ``int` `i ,j;` `        ``// Loop to select Elements one by one` `        ``for` `( i = ``0``; i < size; i++) ` `        ``{` `        `  `            ``// Loop to print position of element` `            ``// where we find an element` `            ``for` `( j = ``0``; j < size; j++)` `            ``{` `        `  `                ``// checking the element in ` `                ``// increasing order` `                ``if` `(arr[j] == i + ``1``) ` `                ``{` `                    ``// print position of element` `                    ``// where element is in inverse` `                    ``// permutation` `                    ``System.out.print( j + ``1` `+ ``" "``);` `                    ``break``;` `                ``}` `            ``}` `        ``}` `    ``}` `    `  `    ``// Driver program to test above function` `    `  `    `  `    ``public` `static` `void` `main (String[] args)` `    ``{` `        ``int` `arr[] = {``2``, ``3``, ``4``, ``5``, ``1``};` `        ``int` `size = arr.length;` `        ``inversePermutation(arr, size);` `        `  `    ``}` `}`   `// This code is contributed by vt_m`

## Python3

 `# Naive Python3 Program to` `# find inverse permutation.`   `# Function to find inverse permutations` `def` `inversePermutation(arr, size): `   `    ``# Loop to select Elements one by one` `    ``for` `i ``in` `range``(``0``, size): `   `        ``# Loop to print position of element` `        ``# where we find an element` `        ``for` `j ``in` `range``(``0``, size): `   `        ``# checking the element in increasing order` `            ``if` `(arr[j] ``=``=` `i ``+` `1``): `   `                ``# print position of element where` `                ``# element is in inverse permutation` `                ``print``(j ``+` `1``, end ``=` `" "``) ` `                ``break`   `# Driver Code` `arr ``=` `[``2``, ``3``, ``4``, ``5``, ``1``] ` `size ``=` `len``(arr)`   `inversePermutation(arr, size) `   `#This code is contributed by Smitha Dinesh Semwal`

## C#

 `// Naive C# Program to find inverse permutation.` `using` `System;`   `class` `GFG {`   `    ``// java function to find inverse permutations` `    ``static` `void` `inversePermutation(``int` `[]arr, ``int` `size)` `    ``{` `        ``int` `i ,j;` `        ``// Loop to select Elements one by one` `        ``for` `( i = 0; i < size; i++) ` `        ``{` `        `  `            ``// Loop to print position of element` `            ``// where we find an element` `            ``for` `( j = 0; j < size; j++)` `            ``{` `        `  `                ``// checking the element in ` `                ``// increasing order` `                ``if` `(arr[j] == i + 1) ` `                ``{` `                    ``// print position of element` `                    ``// where element is in inverse` `                    ``// permutation` `                    ``Console.Write( j + 1 + ``" "``);` `                    ``break``;` `                ``}` `            ``}` `        ``}` `    ``}` `    `  `    ``// Driver program to test above function` `    `  `    `  `    ``public` `static` `void` `Main ()` `    ``{` `        ``int` `[]arr = {2, 3, 4, 5, 1};` `        ``int` `size = arr.Length;` `        ``inversePermutation(arr, size);` `        `  `    ``}` `}`   `// This code is contributed by vt_m`

## PHP

 ``

## Javascript

 ``

Output

`5 1 2 3 4 `

Time Complexity: O(n*n)
Auxiliary Space: O(1)

Method 2: The idea is to use another array to store index and element mappings

Implementation:

## C++

 `// Efficient CPP Program to find inverse permutation.` `#include ` `using` `namespace` `std;`   `// C++ function to find inverse permutations` `void` `inversePermutation(``int` `arr[], ``int` `size) {`   `  ``// to store element to index mappings` `  ``int` `arr2[size];`   `  ``// Inserting position at their` `  ``// respective element in second array` `  ``for` `(``int` `i = 0; i < size; i++) ` `    ``arr2[arr[i] - 1] = i + 1; `   `  ``for` `(``int` `i = 0; i < size; i++) ` `    ``cout << arr2[i] << ``" "``;  ` `}`   `// Driver program to test above function` `int` `main() {` `  ``int` `arr[] = {2, 3, 4, 5, 1};` `  ``int` `size = ``sizeof``(arr) / ``sizeof``(arr[0]);` `  ``inversePermutation(arr, size);` `  ``return` `0;` `}`   `// The code is contributed by Nidhi goel.`

## Java

 `// Efficient Java Program to find` `// inverse permutation.` `import` `java.io.*;`   `class` `GFG {` `    `  `// function to find inverse permutations` `static` `void` `inversePermutation(``int` `arr[], ``int` `size) {`   `    ``// to store element to index mappings` `    ``int` `arr2[] = ``new` `int``[size];`   `    ``// Inserting position at their` `    ``// respective element in second array` `    ``for` `(``int` `i = ``0``; i < size; i++)` `    ``arr2[arr[i] - ``1``] = i + ``1``;`   `    ``for` `(``int` `i = ``0``; i < size; i++)` `    ``System.out.print(arr2[i] + ``" "``);` `}`   `// Driver program to test above function` `public` `static` `void` `main(String args[]) {` `    ``int` `arr[] = {``2``, ``3``, ``4``, ``5``, ``1``};` `    ``int` `size = arr.length;` `    ``inversePermutation(arr, size);` `}` `}`   `// This code is contributed by Nikita Tiwari.`

## Python3

 `# Efficient Python 3 Program to find` `# inverse permutation.`   `# function to find inverse permutations` `def` `inversePermutation(arr, size) :`   `    ``# To store element to index mappings` `    ``arr2 ``=` `[``0``] ``*``(size)` `    `  `    ``# Inserting position at their` `    ``# respective element in second array` `    ``for` `i ``in` `range``(``0``, size) :` `        ``arr2[arr[i] ``-` `1``] ``=` `i ``+` `1` `    `  `    ``for` `i ``in` `range``(``0``, size) :` `        ``print``( arr2[i], end ``=` `" "``) ` `    `  `# Driver program` `arr ``=` `[``2``, ``3``, ``4``, ``5``, ``1``]` `size ``=` `len``(arr)`   `inversePermutation(arr, size)`   `# This code is contributed by Nikita Tiwari.`

## C#

 `// Efficient C# Program to find` `// inverse permutation.` `using` `System;`   `class` `GFG {` `    `  `// function to find inverse permutations` `static` `void` `inversePermutation(``int` `[]arr, ``int` `size) {`   `    ``// to store element to index mappings` `    ``int` `[]arr2 = ``new` `int``[size];`   `    ``// Inserting position at their` `    ``// respective element in second array` `    ``for` `(``int` `i = 0; i < size; i++)` `    ``arr2[arr[i] - 1] = i + 1;`   `    ``for` `(``int` `i = 0; i < size; i++)` `    ``Console.Write(arr2[i] + ``" "``);` `}`   `// Driver program to test above function` `public` `static` `void` `Main() {` `    ``int` `[]arr = {2, 3, 4, 5, 1};` `    ``int` `size = arr.Length;` `    ``inversePermutation(arr, size);` `}` `}`   `// This code is contributed by vt_m.`

## Javascript

 `// function to find inverse permutations` `function` `inversePermutation(arr, size) {`   `    ``// to store element to index mappings` `    ``let arr2 = [];`   `    ``// Inserting position at their` `    ``// respective element in second array` `    ``for` `(let i = 0; i < size; i++)` `    ``arr2[arr[i] - 1] = i + 1;`   `    ``for` `(let i = 0; i < size; i++)` `    ``console.log(arr2[i] + ``" "``);` `}`   `    ``// Driver program to test above function` `    ``let arr = [2, 3, 4, 5, 1];` `    ``let size = arr.length;` `    ``inversePermutation(arr, size);` `    `  `    ``// This code is contributed by aadityaburujwale.`

Output

`5 1 2 3 4 `

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

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