# Check whether (i,j) exists such that arr[i] != arr[j] and arr[arr[i]] is equal to arr[arr[j]]

• Difficulty Level : Basic
• Last Updated : 14 Apr, 2021

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:

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```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```

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

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## Javascript

 ``
Output:
`Yes`

Time Complexity: O(N2)

Auxiliary Space: O(1)

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