Given an undirected graph with V vertices and E edges and a node X, The task is to find the level of node X in the undirected graph. If X does not exist in the graph then return -1.
Note: Traverse the graph starting from vertex 0.
Examples:
Input: V = 5, Edges = {{0, 1}, {0, 2}, {1, 3}, {2, 4}}, X = 3
Output: 2
Explanation: Starting from vertex 0, at level 0 we have node 0, at level 1 we have nodes 1 and 2 and at level 2 we have nodes 3 and 4. So the answer is 2
The example graph
Input: V = 5, Edges = {{0, 1}, {0, 2}, {1, 3}, {2, 4}}, X = 5
Output: -1
Explanation: Vertex 5 is not present in the given graph so answer is -1An example graph
Approach: The problem can be solved based on the following idea:
The given problem can be solved with the help of level order traversal. We can perform a BFS traversal in order to find the level of the given vertex
Follow the steps mentioned below to implement the idea:
- Find the maximum vertex of the graph and store it in a variable (say maxVertex).
- create adjacency list adj[] of size maxVertex+1.
- Check if X is present or not, if not then return -1.
- To traverse the graph, create a queue for level order traversal.
- Push vertex 0 in a queue, and set a counter level to 0.
- Create a visited array of size maxVertex+1 to mark the visited nodes.
- Start BFS traversal if the value of X is found in front of the queue then return the level.
- Keep popping nodes from the queue till it becomes empty and increment the counter level
- In one iteration, push all the unvisited nodes in the queue connected with the current node
- Increment the level variable to jump to the next level.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to find the level of the given nodeint findLevel(int N, vector<vector<int> >& edges, int X){ // Variable to store maximum vertex of graph int maxVertex = 0; for (auto it : edges) { maxVertex = max({ maxVertex, it[0], it[1] }); } // Creating adjacency list vector<int> adj[maxVertex + 1]; for (int i = 0; i < edges.size(); i++) { adj[edges[i][0]].push_back(edges[i][1]); adj[edges[i][1]].push_back(edges[i][0]); } // If X is not present then return -1 if (X > maxVertex || adj[X].size() == 0) return -1; // Initialize a Queue for BFS traversal queue<int> q; q.push(0); int level = 0; // Visited array to mark the already visited nodes vector<int> visited(maxVertex + 1, 0); visited[0] = 1; // BFS traversal while (!q.empty()) { int sz = q.size(); while (sz--) { auto currentNode = q.front(); q.pop(); if (currentNode == X) { return level; } for (auto it : adj[currentNode]) { if (!visited[it]) { q.push(it); visited[it] = 1; } } } level++; } return -1;}// Driver Codeint main(){ int V = 5; vector<vector<int> > edges = { { 0, 1 }, { 0, 2 }, { 1, 3 }, { 2, 4 } }; int X = 3; // Function call int level = findLevel(V, edges, X); cout << level << endl; return 0;} |
Java
// Java code to implement the approachimport java.util.*;class GFG { // Driver code public static void main(String[] args) { int V = 5; int[][] edges = { { 0, 1 }, { 0, 2 }, { 1, 3 }, { 2, 4 } }; int X = 3; // Function call int level = findLevel(V, edges, X); System.out.println(level); } // Function to find the level of the node public static int findLevel(int N, int[][] edges, int X) { int maxVertex = 0; for (int[] it : edges) { maxVertex = Math.max(maxVertex, Math.max(it[0], it[1])); } // Creating adjacency list ArrayList<Integer>[] adj = new ArrayList[maxVertex + 1]; for (int i = 0; i <= maxVertex; i++) adj[i] = new ArrayList<>(); for (int i = 0; i < edges.length; i++) { adj[edges[i][0]].add(edges[i][1]); adj[edges[i][1]].add(edges[i][0]); } // X is not present if (X > maxVertex || adj[X].size() == 0) return -1; // Queue for BFS traversal LinkedList<Integer> q = new LinkedList<>(); q.offer(0); int level = 0; boolean[] visited = new boolean[maxVertex + 1]; // BFS traversal while (!q.isEmpty()) { int sz = q.size(); while (sz-- > 0) { int currentNode = q.poll(); if (currentNode == X) return level; for (int it : adj[currentNode]) { if (!visited[it]) { q.offer(it); visited[it] = true; } } } level++; } return -1; }} |
Python3
# Python code to implement the approach# Function to find the level of the given nodedef findLevel(N,edges,X): # Variable to store maximum vertex of graph maxVertex=0 for it in edges: maxVertex=max(maxVertex,max(it[0],it[1])) # Creating adjacency list adj=[[] for j in range(maxVertex+1)] for i in range(len(edges)): adj[edges[i][0]].append(edges[i][1]) adj[edges[i][1]].append(edges[i][0]) # If X is not present then return -1 if(X>maxVertex or len(adj[X])==0): return -1 # Initialize a Queue for BFS traversal q=[] q.append(0) level=0 # Visited array to mark the already visited nodes visited=[0]*(maxVertex+1) visited[0]=1 # BFS traversal while(len(q)>0): sz=len(q) while(sz>0): currentNode=q[0] q.pop(0) if(currentNode==X): return level for it in adj[currentNode]: if(not visited[it]): q.append(it) visited[it]=1 sz=sz-1 level=level+1 return -1# Driver CodeV=5edges=[[0,1],[0,2],[1,3],[2,4]]X=3#Function calllevel=findLevel(V,edges,X)print(level)# This code is contributed by Pushpesh Raj. |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;using System.Collections;class GFG { // Function to find the level of the given node static int findLevel(int N, int[, ] edges, int X) { // Variable to store maximum vertex of graph int maxVertex = 0; for (int i = 0; i < edges.GetLength(0); i++) { maxVertex = Math.Max( maxVertex, Math.Max(edges[i, 0], edges[i, 1])); } // Creating adjacency list List<int>[] adj = new List<int>[ maxVertex + 1 ]; for (int i = 0; i <= maxVertex; i++) adj[i] = new List<int>(); for (int i = 0; i < edges.GetLength(0); i++) { adj[edges[i, 0]].Add(edges[i, 1]); adj[edges[i, 1]].Add(edges[i, 0]); } // If X is not present then return -1 if (X > maxVertex || adj[X].Count == 0) return -1; // Initialize a Queue for BFS traversal Queue q = new Queue(); q.Enqueue(0); int level = 0; // Visited array to mark the already visited nodes int[] visited = new int[maxVertex + 1]; for (int i = 0; i <= maxVertex; i++) visited[i] = 0; visited[0] = 1; // BFS traversal while (q.Count != 0) { int sz = q.Count; while (sz-- != 0) { int currentNode = (int)q.Dequeue(); if (currentNode == X) { return level; } for (int i = 0; i < adj[currentNode].Count; i++) { int it = adj[currentNode][i]; if (visited[it] == 0) { q.Enqueue(it); visited[it] = 1; } } } level++; } return -1; } static void Main() { int V = 5; int[, ] edges = new int[, ] { { 0, 1 }, { 0, 2 }, { 1, 3 }, { 2, 4 } }; int X = 3; // Function call int level = findLevel(V, edges, X); Console.Write(level); }}// This code is contributed by garg28harsh. |
Javascript
// JS code to implement the approach// Function to find the level of the given nodefunction findLevel( N, edges, X){ // Variable to store maximum vertex of graph let maxVertex = 0; for (let i=0;i<edges.length;i++){ let it = edges[i]; let a = Math.max(it[0],it[1]); maxVertex = Math.max(maxVertex,a); } // Creating adjacency list let adj = []; for(let i=0;i<maxVertex+1;i++){ adj.push([]); } for (let i = 0; i < edges.length; i++) { adj[edges[i][0]].push(edges[i][1]); adj[edges[i][1]].push(edges[i][0]); } // If X is not present then return -1 if (X > maxVertex || adj[X].length == 0) return -1; // Initialize a Queue for BFS traversal let q = []; q.push(0); let level = 0; // Visited array to mark the already visited nodes let visited = []; for(let i=0;i<maxVertex+1;i++) { visited.push(0); } visited[0] = 1; // BFS traversal while (q.length > 0) { let sz = q.length; while (sz--) { let currentNode = q[0]; q.shift(); if (currentNode == X) { return level; } for(let k =0;k<adj[currentNode].length;k++){ let it = adj[currentNode][k]; if (visited[it]==0) { q.push(it); visited[it] = 1; } } } level++; } return -1;}// Driver Codelet V = 5;let edges = [ [ 0, 1 ], [ 0, 2 ], [ 1, 3 ], [ 2, 4 ] ];let X = 3;// Function calllet level = findLevel(V, edges, X);console.log(level);// This code is contributed by ksam24000 |
Output
2
Time Complexity: O(V + E) For traversing all of the nodes.
Auxiliary Space: O(V) to store all the nodes in the queue.
Related Articles:
- Introduction to Graphs – Data Structure and Algorithm Tutorials
- Breadth First Search or BFS for a Graph
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