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:-

1. A[i] is not equal to A[j].
2. 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:

1. Start traversing the Array Arr[] by running two loops.
2. The variable i point at the index 0 and variable j point to the next of i.
3. 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.
4. Repeat the above step till all the elements/index gets traversed.
5. 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)