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# Count of pairs having each element equal to index of the other from an Array

• Difficulty Level : Easy
• Last Updated : 19 Mar, 2021

Given an integer N and an array arr[] that contains elements in the range [1, N], the task is to find the count of all pairs (arr[i], arr[j]) such that i < j and i == arr[j] and j == arr[i].
Examples:

Input: N = 4, arr[] = {2, 1, 4, 3}
Output:
Explanation:
All possible pairs are {1, 2} and {3, 4}
Input: N = 5, arr[] = {5, 5, 5, 5, 1}
Output:
Explanation:
Only possible pair: {1, 5}

Naive Approach:
The simplest approach is to generate all possible pairs of the given array and if any pair satisfies the given condition, increase count. Finally, print the value of count
Time Complexity: O(N2
Auxiliary Space: O(1)
Efficient Approach:
Follow the steps below to solve the above approach:

• Traverse the given array and keep the count of elements(say cnt) whose index equals to arr[arr[index] – 1] – 1. This will count the valid pair with the given criteria.
• After traversal, the total count is given by cnt/2 as we have count every pair twice in the above traversal.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach``#include``using` `namespace` `std;` `// Function to print the count of pair``void` `countPairs(``int` `N, ``int` `arr[])``{``    ``int` `count = 0;` `    ``// Iterate over all the``    ``// elements of the array``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``if` `(i == arr[arr[i] - 1] - 1)``        ``{``            ` `            ``// Increment the count``            ``count++;``        ``}``    ``}` `    ``// Print the result``    ``cout << (count / 2) << endl;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 1, 4, 3 };``    ``int` `N = ``sizeof``(arr)/``sizeof``(arr);` `    ``countPairs(N, arr);``}` `// This code is contributed by Amit Katiyar`

## Java

 `// Java Program to implement``// the above approach``import` `java.util.*;` `class` `GFG {` `    ``// Function to print the count of pair``    ``static` `void` `countPairs(``int` `N, ``int``[] arr)``    ``{``        ``int` `count = ``0``;` `        ``// Iterate over all the``        ``// elements of the array``        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``if` `(i == arr[arr[i] - ``1``] - ``1``) {` `                ``// Increment the count``                ``count++;``            ``}``        ``}` `        ``// Print the result``        ``System.out.println(count / ``2``);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``2``, ``1``, ``4``, ``3` `};``        ``int` `N = arr.length;` `        ``countPairs(N, arr);``    ``}``}`

## Python3

 `# Python3 program to implement``# the above approach``# Function to print the count of pair``def` `countPairs(N, arr):` `    ``count ``=` `0` `    ``# Iterate over all the``    ``# elements of the array``    ``for` `i ``in` `range``(N):``        ``if` `(i ``=``=` `arr[arr[i] ``-` `1``] ``-` `1``):``       ` `            ``# Increment the count``            ``count ``+``=` `1` `    ``# Print the result``    ``print``(count ``/``/` `2``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``  ` `    ``arr ``=` `[``2``, ``1``, ``4``, ``3``]``    ``N ``=` `len``(arr)``    ``countPairs(N, arr)` `# This code is contributed by Chitranayal`

## C#

 `// C# Program to implement``// the above approach``using` `System;``class` `GFG{`` ` `  ``// Function to print the count of pair``  ``static` `void` `countPairs(``int` `N, ``int``[] arr)``  ``{``    ``int` `count = 0;` `    ``// Iterate over all the``    ``// elements of the array``    ``for` `(``int` `i = 0; i < N; i++)``    ``{``      ``if` `(i == arr[arr[i] - 1] - 1)``      ``{` `        ``// Increment the count``        ``count++;``      ``}``    ``}` `    ``// Print the result``    ``Console.Write(count / 2);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args)``  ``{``    ``int``[] arr = { 2, 1, 4, 3 };``    ``int` `N = arr.Length;` `    ``countPairs(N, arr);``  ``}``}` `// This code is contributed by Ritik Bansal`

## Javascript

 ``
Output:

`2`

Time Complexity: O(N)
Auxiliary Space: O(1)

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