Total number of components in the index Array
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: 2
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: 2
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:
- Start from the first unvisited index which will be index with integer 0 in it.
- During DFS Traversal mark the visited elements in the array until the elements form a cycle.
- If a cycle is formed then it means that we have got one component and hence increase the component count.
- Repeat all the above steps for all the unvisited index in the array and count the total components in the given index array.
- 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++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// To mark the visited index// during DFS Traversalvector<int> visited;// Function for DFS traversalvoid dfs(int i, int a[]){ // Check if not visited then // recurr for the next index if (visited[i] == 0) { visited[i] = 1; // DFS Traversal for next index dfs(a[i], a); } return;}// Function for checking which// indexes are remainingint allvisited(int a[], int n){ for (int i = 0; i < n; i++) { // Marks that the ith // index is not visited if (visited[i] == 0) return i; } return -1;}// Function for counting componentsint count(int a[], int n){ int c = 0; // Function call int x = allvisited(a, n); while (x != -1) { // Count number of components c++; // DFS call dfs(x, a); x = allvisited(a, n); } // Print the total count of components cout << c << endl;}// Driver Codeint main(){ // Given array arr[] int arr[] = { 1, 4, 3, 5, 0, 2, 6 }; int N = sizeof(arr) / sizeof(arr[0]); visited = vector<int>(N+1,0); // Function Call count(arr, N); return 0;} |
Java
// Java program for the above approachimport 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 |
Python3
# Python3 program for the above approach # Function for DFS traversaldef 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 remainingdef 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 componentsdef 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 Codeif __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 |
C#
// C# program for the above approachusing System;using System.Collections;using System.Collections.Generic;class GFG{ // Function for DFS traversalstatic 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 remainingstatic 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 componentsstatic 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 Codepublic 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 |
Javascript
<script>// Javascript program for the above approach// Function for DFS traversalfunction dfs(i, a, m){ // Check if not visited then // recurr for the next index if (!m.has(i)) { m.set(i, 1); m[i] = 1; // DFS Traversal for next index dfs(a[i], a, m); } return;}// Function for checking which// indexes are remainingfunction allvisited(a, n, m){ for (var i = 0; i < n; i++) { // Marks that the ith // index is not visited if (!m.has(i)) return i; } return -1;}// Function for counting componentsfunction count(a, n){ var c = 0; // To mark the visited index // during DFS Traversal var m = new Map(); // Function call var 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 document.write( c );}// Driver Code// Given array arr[]var arr = [1, 4, 3, 5, 0, 2, 6];var N = arr.length;// Function Callcount(arr, N);// This code is contributed by itsok.</script> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(N)
