Create connected-components.md

tutorials/graph-theory/connected-components.md

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+---
+title: 'Connected-Components'
+description: 'Author: @adityabisht02'
+hide_table_of_contents: true
+---
+
+<TutorialAuthors names="Aditya Bisht"/>
+
+## Overview
+In a graph, sometimes all nodes might not be connected with each other. Let's take an example of a graph with 5 nodes-
+
+
+![image](https://user-images.githubusercontent.com/89146189/195919108-2e06dbbe-717a-4f86-9ba1-90cbdf5c119c.png)
+
+Here you can see that 1-2-3 are connected and 4-5 are connected.
+Here , 1-2-3 is called one connected component.
+
+## How to determine if a connected component is present ?
+1. Start a BFS or DFS from any node of the graph.
+2. If you are able to visit all the other nodes from that single node that means the all nodes in the graph are connected.
+3. If you are unable to reach a particular node or a group of nodes, then that group of nodes might be a different connected component.
+
+## How to find connected components?
+1. Let's an adjacency list based graph is given.
+2. Traverse the adjacency list and for every node do a BFS or DFS on the node and mark all the nodes you visit in this way as visited.
+3. While traversing the list do not do BFS on the nodes which have already been visited.
+4. Hence everytime you do a BFS/DFS you are traversing a connected component so you can count the number of times you do a BFS/DFS.
+
+
+## Sample question:
+https://leetcode.com/problems/number-of-provinces/
+
+Solution:
+
+class Solution {
+
+    //bfs function
+    public void bfs(int isConnected[][],int visited[],int node){
+
+        Queue<Integer> q=new LinkedList<>();
+        q.offer(node);
+
+        while(!q.isEmpty()){
+            int current=q.poll();
+            visited[current]=1;
+
+           for(int i=0;i<isConnected[0].length;i++){
+               if(isConnected[current][i]!=0 && visited[i]!=1){
+                   q.offer(i);
+                   visited[i]=1;
+               }
+           }
+        }
+
+
+    }
+
+    public int findCircleNum(int[][] isConnected) {
+
+        int i,j,provincecount=0;
+        int visited[]=new int[isConnected.length];
+
+
+        //traverse through the adjacency array
+        for(i=0;i<isConnected.length;i++){
+
+            if(visited[i]==1){
+                continue;
+            }
+            for(j=0;j<isConnected[0].length;j++){
+
+                //if not visited and is connected
+                if(visited[j]==0 && isConnected[i][j]==1){
+                    //everytime u do a bfs u complete one connected component
+                     bfs(isConnected,visited,j);
+                    provincecount++;
+                }
+
+            }
+        }
+
+        return provincecount;
+    }
+}
+
+## Similar problems on leetcode:
+1. Number of Islands - https://leetcode.com/problems/number-of-islands/
+2. Number of Connected Components- https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/
+