Inverse Permutation

• Difficulty Level : Easy
• Last Updated : 27 May, 2021

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.

C++

 // Naive CPP Program to find inverse permutation.#include using namespace std; // C++ function to find inverse permutationsvoid 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 functionint main() {  int arr[] = {2, 3, 4, 5, 1};  int size = sizeof(arr) / sizeof(arr);  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 permutationsdef 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 Codearr = [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



Javascript



Output :

5 1 2 3 4

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

C++

 // Efficient CPP Program to find inverse permutation.#include using namespace std; // C++ function to find inverse permutationsvoid 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 functionint main() {  int arr[] = {2, 3, 4, 5, 1};  int size = sizeof(arr) / sizeof(arr);  inversePermutation(arr, size);  return 0;}

Java

 // Efficient Java Program to find// inverse permutation.import java.io.*; class GFG {     // function to find inverse permutationsstatic 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 functionpublic 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 permutationsdef inversePermutation(arr, size) :     # To store element to index mappings    arr2 =  *(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 programarr = [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 permutationsstatic 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 functionpublic static void Main() {    int []arr = {2, 3, 4, 5, 1};    int size = arr.Length;    inversePermutation(arr, size);}} // This code is contributed by vt_m.
Output : 5 1 2 3 4

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