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 <iostream> 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 arraystatic 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 codepublic 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 coden = 4arr = [ 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 arraystatic 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 codepublic static void Main(String[] args){ int n = 4; int[] arr = { 1, 4, 3, 2 }; inverseEqual(arr, n);}}// This code is contributed by sapnasingh4991 |
Javascript
<script>// Javascript Program to implement // the above approach // Function to check if the inverse // permutation of the given array is // same as the original array function inverseEqual(arr, n) { // Stores the inverse // permutation var brr = Array(n).fill(0); // Generate the inverse permutation for (var i = 0; i < n; i++) { var present_index = arr[i] - 1; brr[present_index] = i + 1; } // Check if the inverse permutation // is same as the given array for (var i = 0; i < n; i++) { if (arr[i] != brr[i]) { document.write( "No" ); return; } } document.write( "Yes" ); } // Driver Code var n = 4; var arr = [ 1, 4, 3, 2 ]; inverseEqual(arr, n); // This code is contributed by noob2000.</script> |
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 <bits/stdc++.h>using namespace std;// Function to check if the inverse// permutation of the given array is// same as the original arraybool 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 Codeint 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<n;i++){ if (arr[arr[i] - 1] != i + 1) return false; } return true; } // Driver code public static void main(String args[]) { int n = 4; int[] arr = { 1, 4, 3, 2 }; // Function Call System.out.println((inverseEqual(arr, n)==true)?"Yes" : "No"); }}// This Code is contributed by shinjanpatra |
Python3
# Python program to implement# the above approach# Function to check if the inverse# permutation of the given array is# same as the original arraydef 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 Coden = 4arr = [ 1, 4, 3, 2 ]# Function Callprint("Yes" if (inverseEqual(arr, n)==True) else "No")# This code is contributed by shinjanpatra |
C#
// C# Program to implement// the above approachusing System;class GFG{// Function to check if the inverse// permutation of the given array is// same as the original arraystatic 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 codepublic static void Main(String[] args){ int n = 4; int[] arr = { 1, 4, 3, 2 }; // Function Call Console.WriteLine((inverseEqual(arr, n)==true)?"Yes" : "No");}}// This Code is contributed by Pushpesh Raj. |
Javascript
<script>// Javascript program to implement// the above approach// Function to check if the inverse// permutation of the given array is// same as the original arrayfunction inverseEqual(arr, n){ // Check the if inverse permutation // is not same for (let i = 0; i < n; i++) if (arr[arr[i] - 1] != i + 1) return false; return true;} // Driver Code let n = 4; let arr = [ 1, 4, 3, 2 ]; // Function Call document.write(inverseEqual(arr, n) ? "Yes" : "No");</script> |
Output
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
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