# 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 != to A but A[A] == A[A]

Input: N = 4, A[] = {2, 1, 3, 3}
Output: No
As A[A] == A[A] but A == A
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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); ` ` `  `    ``// 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

 ` `

Output:

```Yes
```

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