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:
// 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> |
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:
// 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.
|
5 1 2 3 4
Time Complexity: O(n)
Auxiliary Space: O(n)