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

Last Updated : 25 Jan, 2023

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: 2
Explanation:
All possible pairs are {1, 2} and {3, 4}

Input: N = 5, arr[] = {5, 5, 5, 5, 1}
Output: 1
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)

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

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# 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>