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

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  • Difficulty Level : Easy
  • Last Updated : 25 Jan, 2023
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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

C++




// 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 - 1; i++) {
        for (int j = i + 1; j < N; j++) {
            if ((i + 1) == arr[j] && (j + 1) == arr[i])
                count++;
        }
    }
 
    // Print the result
    cout << count << endl;
}
 
// Driver Code
int main()
{
    int arr[] = { 5, 5, 5, 5, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    countPairs(N, arr);
}

Java




import java.util.Arrays;
 
public class Gfg {
    public static void main(String[] args)
    {
        int[] arr = { 5, 5, 5, 5, 1 };
        int N = arr.length;
        int count = 0;
 
        // Iterate over all the
        // elements of the array
        for (int i = 0; i < N - 1; i++) {
            for (int j = i + 1; j < N; j++) {
                if ((i + 1) == arr[j] && (j + 1) == arr[i])
                    count++;
            }
        }
        // Print the result
        System.out.println(count);
    }
}

Python3




def countPairs(N, arr):
    count = 0
    # Iterate over all the
    # elements of the array
    for i in range(N-1):
        for j in range(i+1, N):
            if (i + 1) == arr[j] and (j + 1) == arr[i]:
                count += 1
 
    # Print the result
    print(count)
 
# Driver code
if __name__ == "__main__":
    arr = [5, 5, 5, 5, 1]
    N = len(arr)
    countPairs(N, arr)

C#




using System;
 
class Gfg
{
    public static void Main(string[] args)
    {
        int[] arr = { 5, 5, 5, 5, 1 };
        int N = arr.Length;
        int count = 0;
 
        // Iterate over all the
        // elements of the array
        for (int i = 0; i < N - 1; i++)
        {
            for (int j = i + 1; j < N; j++)
            {
                if ((i + 1) == arr[j] && (j + 1) == arr[i])
                    count++;
            }
        }
        // Print the result
        Console.WriteLine(count);
    }
}

Javascript




// Javascript program to implement
// the above approach
 
// Function to print the count of pair
function countPairs(N, arr)
{
    let count = 0;
 
    // Iterate over all the
    // elements of the array
    for (let i = 0; i < N - 1; i++) {
        for (let j = i + 1; j < N; j++) {
            if ((i + 1) == arr[j] && (j + 1) == arr[i])
                count++;
        }
    }
 
    // Print the result
    console.log(count);
}
 
// Driver Code
let arr = [ 5, 5, 5, 5, 1 ];
let N = arr.length;
 
countPairs(N, arr);
 
// This code is contributed by agrawalpoojaa976.

Output

1

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<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

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




<script>
 
    // Javascript program to implement
    // the above approach
     
    // Function to print the count of pair
    function countPairs(N, arr)
    {
        let count = 0;
 
        // Iterate over all the
        // elements of the array
        for(let i = 0; i < N; i++)
        {
            if (i == arr[arr[i] - 1] - 1)
            {
 
                // Increment the count
                count++;
            }
        }
 
        // Print the result
        document.write(count / 2);
    }
     
    let arr = [ 2, 1, 4, 3 ];
    let N = arr.length;
  
    countPairs(N, arr);
     
</script>

Output

2

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


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