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 <bits/stdc++.h>
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;
}
// 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
# 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
// 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
// Naive PHP Program to
// find inverse permutation.
// Function to find
// inverse permutations
function inversePermutation($arr, $size)
{
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
echo $j + 1 , " ";
break;
}
}
}
}
// Driver Code
$arr = array(2, 3, 4, 5, 1);
$size = sizeof($arr);
inversePermutation($arr, $size);
// This code is contributed by aj_36
?>
<script>
// Naive JavaScript program to find inverse permutation.
// JavaScript function to find inverse permutations
function inversePermutation(arr, size)
{
let 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
document.write( j + 1 + " ");
break;
}
}
}
}
// Driver code
let arr = [2, 3, 4, 5, 1];
let size = arr.length;
inversePermutation(arr, size);
</script>
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 <bits/stdc++.h>
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.
// 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.
# 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.
// 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.
// 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.
Time Complexity: O(n)
Auxiliary Space: O(n)
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