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

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[0]);       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

Output:

2

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

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