Number of Connected Components in an Undirected Graph in C++

C++Server Side ProgrammingProgramming

Suppose we have n nodes and they are labeled from 0 to n - 1 and a list of undirected edges, are also given, we have to define one function to find the number of connected components in an undirected graph.

So, if the input is like n = 5 and edges = [[0, 1], [1, 2], [3, 4]],

then the output will be 2

To solve this, we will follow these steps −

  • Define a function dfs(), this will take node, graph, an array called visited,

  • if visited[node] is false, then −

    • visited[node] := true

  • for initialize i := 0, when i < size of graph[node], update (increase i by 1), do −

    • dfs(graph[node, i], graph, visited)

  • From the main method do the following −

  • Define an array visited of size n

  • if not n is non-zero, then −

    • Define an array graph[n]

  • for initialize i := 0, when i < size of edges, update (increase i by 1), do −

    • u := edges[i, 0]

    • v := edges[i, 1]

    • insert v at the end of graph[u]

    • insert u at the end of graph[v]

  • ret := 0

  • for initialize i := 0, when i < n, update (increase i by 1), do −

    • if not visited[i] is non-zero, then −

      • dfs(i, graph, visited)

      • (increase ret by 1)

  • return ret

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   void dfs(int node, vector<int< graph[], vector<bool>& visited){
      if(visited[node]) return;
         visited[node] = true;
      for(int i = 0; i < graph[node].size(); i++){
         dfs(graph[node][i], graph, visited);
      }
   }
   int countComponents(int n, vector<vector<int<>& edges) {
      vector <bool> visited(n);
      if(!n) return 0;
         vector <int< graph[n];
      for(int i = 0; i < edges.size(); i++){
         int u = edges[i][0];
         int v = edges[i][1];
         graph[u].push_back(v);
         graph[v].push_back(u);
      }
      int ret = 0;
      for(int i = 0; i < n; i++){
         if(!visited[i]){
            dfs(i, graph, visited);
            ret++;
         }
      }
      return ret;
   }
};
main(){
   Solution ob;
   vector<vector<int<> v = {{0,1},{1,2},{3,4}};
   cout << (ob.countComponents(5, v));
}

Input

5, [[0,1],[1,2],[3,4]]

Output

2
raja
Published on 18-Nov-2020 12:13:25
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