Check if the given array is mirror-inverse
Given an array arr[], the task is to find whether the array is mirror inverse. Inverse of an array means if the array elements are swapped with their corresponding indices and the array is called mirror-inverse if it’s inverse is equal to itself. If array is mirror-inverse then print Yes else print No.
Examples:
Input: arr[] = [3, 4, 2, 0, 1}
Output: Yes
Explanation:
In the given array:
index(0) -> value(3)
index(1) -> value(4)
index(2) -> value(2)
index(3) -> value(0)
index(4) -> value(1)
To find the inverse of the array, swap the index and the value of the array.
index(3) -> value(0)
index(4) -> value(1)
index(2) -> value(2)
index(0) -> value(3)
index(1) -> value(4)
Inverse arr[] = {3, 4, 2, 0, 1}
So, the inverse array is equal to the given array.Input: arr[] = {1, 2, 3, 0}
Output: No
A simple approach is to create a new array by swapping the value and index of the given array and check whether the new array is equal to the original array or not.
A better approach is to traverse the array and for all the indices, if arr[arr[index]] = index is satisfied then the given array is mirror inverse.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include<bits/stdc++.h> using namespace std; // Function that returns true if // the array is mirror-inverse bool isMirrorInverse( int arr[], int n) { for ( int i = 0; i < n; i++) { // If condition fails for any element if (arr[arr[i]] != i) return false ; } // Given array is mirror-inverse return true ; } // Driver code int main() { int arr[] = { 1, 2, 3, 0 }; int n = sizeof (arr)/ sizeof (arr[0]); if (isMirrorInverse(arr,n)) cout << "Yes" ; else cout << "No" ; return 0; } // This code is contributed by Rajput-Ji |
Java
// Java implementation of the approach public class GFG { // Function that returns true if // the array is mirror-inverse static boolean isMirrorInverse( int arr[]) { for ( int i = 0 ; i < arr.length; i++) { // If condition fails for any element if (arr[arr[i]] != i) return false ; } // Given array is mirror-inverse return true ; } // Driver code public static void main(String[] args) { int arr[] = { 1 , 2 , 3 , 0 }; if (isMirrorInverse(arr)) System.out.println( "Yes" ); else System.out.println( "No" ); } } |
Python3
# Python 3 implementation of the approach # Function that returns true if # the array is mirror-inverse def isMirrorInverse(arr, n) : for i in range (n) : # If condition fails for any element if (arr[arr[i]] ! = i) : return False ; # Given array is mirror-inverse return true; # Driver code if __name__ = = "__main__" : arr = [ 1 , 2 , 3 , 0 ]; n = len (arr) ; if (isMirrorInverse(arr,n)) : print ( "Yes" ); else : print ( "No" ); # This code is contributed by Ryuga |
C#
// C# implementation of the approach using System; class GFG { // Function that returns true if // the array is mirror-inverse static bool isMirrorInverse( int []arr) { for ( int i = 0; i < arr.Length; i++) { // If condition fails for any element if (arr[arr[i]] != i) return false ; } // Given array is mirror-inverse return true ; } // Driver code static public void Main () { int []arr = { 1, 2, 3, 0 }; if (isMirrorInverse(arr)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by ajit... |
PHP
<?php // PHP implementation of the approach // Function that returns true if // the array is mirror-inverse function isMirrorInverse( $arr ) { for ( $i = 0; $i < sizeof( $arr ); $i ++) { // If condition fails for any element if ( $arr [ $arr [ $i ]] != $i ) return false; } // Given array is mirror-inverse return true; } // Driver code $arr = array (1, 2, 3, 0); if (isMirrorInverse( $arr )) echo ( "Yes" ); else echo ( "No" ); // These code is contributed by Code_Mech. ?> |
Javascript
<script> // JavaScript implementation of the approach // Function that returns true if // the array is mirror-inverse function isMirrorInverse(arr) { for (i = 0; i < arr.length; i++) { // If condition fails for any element if (arr[arr[i]] != i) return false ; } // Given array is mirror-inverse return true ; } // Driver code var arr = [ 1, 2, 3, 0 ]; if (isMirrorInverse(arr)) document.write( "Yes" ); else document.write( "No" ); // This code contributed by Rajput-Ji </script> |
No
Time complexity: O(N), where N is the size of the given array.
Auxiliary space: O(1), no extra space is required.
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