Check whether (i,j) exists such that arr[i] != arr[j] and arr[arr[i]] is equal to arr[arr[j]]
Given an array A[]. The task is to determine if it is possible to choose two indices ‘i’ and ‘j’ such that the below conditions gets satisfied:-
- A[i] is not equal to A[j].
- A[A[i]] is equal to A[A[j]].
Note: The value of the elements in an array is less than the value of N i.e. For every i, arr[i] < N.
Examples:
Input: N = 4, A[] = {1, 1, 2, 3} Output: Yes As A[3] != to A[1] but A[A[3]] == A[A[1]] Input: N = 4, A[] = {2, 1, 3, 3} Output: No As A[A[3]] == A[A[4]] but A[3] == A[4]
Approach:
- Start traversing the Array Arr[] by running two loops.
- The variable i point at the index 0 and variable j point to the next of i.
- If Arr[i] is not equal to Arr[j] then check if Arr[Arr[i] – 1] is equal to Arr[Arr[j] – 1]. If yes then return true.
Else check Arr[Arr[i]- 1] and Arr[Arr[j] – 1] for other indices also. - Repeat the above step till all the elements/index gets traversed.
- If no such indices found return false.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function that will tell whether // such Indices present or Not. bool checkIndices( int Arr[], int N) { for ( int i = 0; i < N - 1; i++) { for ( int j = i + 1; j < N; j++) { // Checking 1st condition i.e whether // Arr[i] equal to Arr[j] or not if (Arr[i] != Arr[j]) { // Checking 2nd condition i.e whether // Arr[Arr[i]] equal to Arr[Arr[j]] or not. if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1]) return true ; } } } return false ; } // Driver Code int main() { int Arr[] = { 3, 2, 1, 1, 4 }; int N = sizeof (Arr) / sizeof (Arr[0]); // Calling function. checkIndices(Arr, N) ? cout << "Yes" : cout << "No" ; return 0; } |
Java
// Java implementation of the above approach // Function that calculates marks. class GFG { static boolean checkIndices( int Arr[], int N) { for ( int i = 0 ; i < N - 1 ; i++) { for ( int j = i + 1 ; j < N; j++) { // Checking 1st condition i.e whether // Arr[i] equal to Arr[j] or not if (Arr[i] != Arr[j]) { // Checking 2nd condition i.e whether // Arr[Arr[i]] equal to Arr[Arr[j]] or not. if (Arr[Arr[i] - 1 ] == Arr[Arr[j] - 1 ]) return true ; } } } return false ; } // Driver code public static void main(String args[]) { int Arr[] = { 3 , 2 , 1 , 1 , 4 }; int N = Arr.length; if (checkIndices(Arr, N)) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is Contributed by // Naman_Garg |
Python 3
# Python 3 implementation of the # above approach # Function that will tell whether # such Indices present or Not. def checkIndices(Arr, N): for i in range (N - 1 ): for j in range (i + 1 , N): # Checking 1st condition i.e whether # Arr[i] equal to Arr[j] or not if (Arr[i] ! = Arr[j]): # Checking 2nd condition i.e whether # Arr[Arr[i]] equal to Arr[Arr[j]] or not. if (Arr[Arr[i] - 1 ] = = Arr[Arr[j] - 1 ]): return True return False # Driver Code if __name__ = = "__main__" : Arr = [ 3 , 2 , 1 , 1 , 4 ] N = len (Arr) # Calling function. if checkIndices(Arr, N): print ( "Yes" ) else : print ( "No" ) # This code is contributed by ita_c |
C#
// C# implementation of the above approach using System; class GFG { // Function that calculates marks. static bool checkIndices( int []Arr, int N) { for ( int i = 0; i < N - 1; i++) { for ( int j = i + 1; j < N; j++) { // Checking 1st condition i.e whether // Arr[i] equal to Arr[j] or not if (Arr[i] != Arr[j]) { // Checking 2nd condition i.e // whether Arr[Arr[i]] equal // to Arr[Arr[j]] or not. if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1]) return true ; } } } return false ; } // Driver code static public void Main () { int []Arr = { 3, 2, 1, 1, 4 }; int N = Arr.Length; if (checkIndices(Arr, N)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is Contributed by Sachin |
PHP
<?php // PHP implementation of the // above approach // Function that will tell whether // such Indices present or Not. function checkIndices( $Arr , $N ) { for ( $i = 0; $i < $N - 1; $i ++) { for ( $j = $i + 1; $j < $N ; $j ++) { // Checking 1st condition i.e // whether Arr[i] equal to // Arr[j] or not if ( $Arr [ $i ] != $Arr [ $j ]) { // Checking 2nd condition i.e // whether Arr[Arr[i]] equal to // Arr[Arr[j]] or not. if ( $Arr [ $Arr [ $i ] - 1] == $Arr [ $Arr [ $j ] - 1]) return true; } } } return false; } // Driver Code $Arr = array (3, 2, 1, 1, 4); $N = sizeof( $Arr ); // Calling function. if (checkIndices( $Arr , $N )) echo "Yes" ; else echo "No" ; // This code is contributed // by Akanksha Rai ?> |
Javascript
<script> // Javascript implementation of the above approach // Function that will tell whether // such Indices present or Not. function checkIndices(Arr, N) { for ( var i = 0; i < N - 1; i++) { for ( var j = i + 1; j < N; j++) { // Checking 1st condition i.e whether // Arr[i] equal to Arr[j] or not if (Arr[i] != Arr[j]) { // Checking 2nd condition i.e whether // Arr[Arr[i]] equal to Arr[Arr[j]] or not. if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1]) return true ; } } } return false ; } // Driver Code var Arr = [ 3, 2, 1, 1, 4 ]; var N = Arr.length; // Calling function. checkIndices(Arr, N) ? document.write( "Yes" ) : document.write( "No" ); </script> |
Output
Yes
Complexity Analysis:
- Time Complexity: O(N2)
- Auxiliary Space: O(1)
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