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++

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// C++ program to implement
// the above approach
#include<bits/stdc++.h>
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

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Java

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// 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);
    }
}

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Python3

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# 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

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C#

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// 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

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Output: 

2



 

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

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