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Total number of components in the index Array
  • Difficulty Level : Expert
  • Last Updated : 18 Dec, 2020

Given an array arr[] of N integers of value from 0 to N, the task is to count the number of components in Index Array. 

Index array means if we are at ith index then it leads to arr[i]. 
The component of an index array is counted when it forms a cycle. If no cycle persists or the array contains a single element then also we consider it as a component. 
For Example: 
Let array arr[] = {1, 2, 0, 3} 
{1, 2, 0} will form one component as starting from index 0 we reach the index 0 again as: 
1 -> 2(arr[1]) -> 0(arr[2]) -> 1(arr[0]) 
 

Examples: 

Input: arr[] = {1, 2, 3, 5, 0, 4, 6} 
Output:
Explanation: 
Below is the traversal of the 2 components: 
Component 1: Start traversal from 0, then the path of traversal is given by: 
1 -> 2(arr[1]) -> 3(arr[2]) -> 5(arr[3]) -> 4(arr[5]) -> 0(arr[4]) -> 1(arr[0]). 
Component 2: Only 6 is unvisited it creates one more component. 
So, the total components = 2.

Input: arr[] = {1, 2, 0, 3} 
Output:
Explanation: 
Below is the traversal of the 2 components: 
Component 1: Start traversal from 0, then the path of traversal is given by: 
1 -> 2(arr[1]) -> 0(arr[2]) -> 1(arr[0]) 
Component 2: Only 3 is unvisited it creates one more component. 
So, the total components = 2. 



Approach: The idea is to use the concept of DFS traversal. Below are the steps:  

  1. Start from the first unvisited index which will be index with integer 0 in it.
  2. During DFS Traversal mark the visited elements in the array until the elements form a cycle.
  3. If a cycle is formed then it means that we have got one component and hence increase the component count.
  4. Repeat all the above steps for all the unvisited index in the array and count the total components in the given index array.
  5. If all the index of the array are visited, then print the total count of connected components.

Below is the implementation of the above approach: 

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function for DFS traversal
void dfs(int i, int a[],
         map<int, int>& m)
{
    // Check if not visited then
    // recurr for the next index
    if (m[i] == 0) {
        m[i] = 1;
 
        // DFS Traversal for next index
        dfs(a[i], a, m);
    }
 
    return;
}
 
// Function for checking which
// indexes are remaining
int allvisited(int a[], int n,
               map<int, int>& m)
{
    for (int i = 0; i < n; i++) {
 
        // Marks that the ith
        // index is not visited
        if (m[i] == 0)
            return i;
    }
    return -1;
}
 
// Function for counting components
int count(int a[], int n)
{
    int c = 0;
 
    // To mark the visited index
    // during DFS Traversal
    map<int, int> m;
 
    // Function call
    int x = allvisited(a, n, m);
 
    while (x != -1) {
 
        // Count number of components
        c++;
 
        // DFS call
        dfs(x, a, m);
 
        x = allvisited(a, n, m);
    }
 
    // Print the total count of components
    cout << c << endl;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 4, 3, 5, 0, 2, 6 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    count(arr, N);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
class Main
{
    // Function for DFS traversal
    static void dfs(int i, int[] a,
                    HashMap<Integer, Integer> m)
    {
          
        // Check if not visited then
        // recurr for the next index
        if (!m.containsKey(i))
        {
            m.put(i, 1);
              
            // DFS Traversal for next index
            dfs(a[i], a, m);
        }
        return;
    }
       
    // Function for checking which
    // indexes are remaining
    static int allvisited(int[] a, int n,
                          HashMap<Integer, Integer> m)
    {
        for(int i = 0; i < n; i++)
        {
              
            // Marks that the ith
            // index is not visited
            if (!m.containsKey(i))
                return i;
        }
        return -1;
    }
       
    // Function for counting components
    static void count(int[] a, int n)
    {
        int c = 0;
          
        // To mark the visited index
        // during DFS Traversal
        HashMap<Integer, Integer> m = new HashMap<>();
                                             
        // Function call
        int x = allvisited(a, n, m);
       
        while (x != -1)
        {
              
            // Count number of components
            c++;
              
            // DFS call
            dfs(x, a, m);
       
            x = allvisited(a, n, m);
        }
          
        // Print the total count of components
        System.out.print(c);
    }
 
    public static void main(String[] args)
    {
       
        // Given array arr[]
        int[] arr = { 1, 4, 3, 5, 0, 2, 6 };
       
        int N = arr.length;
       
        // Function Call
        count(arr, N);
    }
}
 
// This code is contributed by divyesh072019

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Python3

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# Python3 program for the above approach
  
# Function for DFS traversal
def dfs(i, a, m):
 
    # Check if not visited then
    # recurr for the next index
    if i in m:
        if m[i] == 0:
            m[i] = 1
             
            # DFS Traversal for next index
            dfs(a[i], a, m)
    else:
        m[i] = 1
         
        # DFS Traversal for next index
        dfs(a[i], a, m)
  
    return
 
# Function for checking which
# indexes are remaining
def allvisited(a, n, m):
     
    for i in range(n):
  
        # Marks that the ith
        # index is not visited
        if i in m:
            if m[i] == 0:
                return i
         
        else:
            return i
 
    return -1
 
# Function for counting components
def count(a, n):
 
    c = 0
  
    # To mark the visited index
    # during DFS Traversal
    m = dict()
  
    # Function call
    x = allvisited(a, n, m)
  
    while (x != -1):
  
        # Count number of components
        c += 1
  
        # DFS call
        dfs(x, a, m)
  
        x = allvisited(a, n, m)
     
    # Print the total count of components
    print(c)
     
# Driver Code
if __name__=='__main__':
 
    # Given array arr[]
    arr = [ 1, 4, 3, 5, 0, 2, 6 ]
     
    N = len(arr)
  
    # Function Call
    count(arr, N)
  
# This code is contributed by rutvik_56

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C#

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// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
  
// Function for DFS traversal
static void dfs(int i, int []a,
     Dictionary<int, int> m)
{
     
    // Check if not visited then
    // recurr for the next index
    if (!m.ContainsKey(i))
    {
        m[i] = 1;
         
        // DFS Traversal for next index
        dfs(a[i], a, m);
    }
    return;
}
  
// Function for checking which
// indexes are remaining
static int allvisited(int []a, int n,
               Dictionary<int, int> m)
{
    for(int i = 0; i < n; i++)
    {
         
        // Marks that the ith
        // index is not visited
        if (!m.ContainsKey(i))
            return i;
    }
    return -1;
}
  
// Function for counting components
static void count(int []a, int n)
{
    int c = 0;
     
    // To mark the visited index
    // during DFS Traversal
    Dictionary<int,
               int> m = new Dictionary<int,
                                       int>();
                                        
    // Function call
    int x = allvisited(a, n, m);
  
    while (x != -1)
    {
         
        // Count number of components
        c++;
         
        // DFS call
        dfs(x, a, m);
  
        x = allvisited(a, n, m);
    }
     
    // Print the total count of components
    Console.Write(c);
}
  
// Driver Code
public static void Main()
{
     
    // Given array arr[]
    int []arr = { 1, 4, 3, 5, 0, 2, 6 };
  
    int N = arr.Length;
  
    // Function Call
    count(arr, N);
}
}
 
// This code is contributed by pratham76

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Output: 

3

 

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

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