Given a connected graph with N vertices and M edges. The task is to print the lexicographically smallest BFS traversal of the graph starting from 1.
Note: The vertices are numbered from 1 to N.
Examples:
Input: N = 5, M = 5
Edges:
1 4
3 4
5 4
3 2
1 5
Output: 1 4 3 2 5
Start from 1, go to 4, then to 3 and then to 2 and to 5.
Input: N = 3, M = 2
Edges:
1 2
1 3
Output: 1 2 3
Approach: Instead of doing a normal BFS traversal on the graph, we can use a priority queue(min-heap) instead of a simple queue. When a node is visited add its adjacent nodes into the priority queue. Every time, we visit a new node, it will be the one with the smallest index in the priority queue. Print the nodes when every time we visit them starting from 1.
Below is the implementation of the above approach:
CPP
// C++ program to print the lexcicographically// smallest path starting from 1#include <bits/stdc++.h>using namespace std;// Function to print the smallest lexicographically// BFS path starting from 1void printLexoSmall(vector<int> adj[], int n){ // Visited array bool vis[n + 1]; memset(vis, 0, sizeof vis); // Minimum Heap priority_queue<int, vector<int>, greater<int> > Q; // First one visited vis[1] = true; Q.push(1); // Iterate till all nodes are visited while (!Q.empty()) { // Get the top element int now = Q.top(); // Pop the element Q.pop(); // Print the current node cout << now << " "; // Find adjacent nodes for (auto p : adj[now]) { // If not visited if (!vis[p]) { // Push Q.push(p); // Mark as visited vis[p] = true; } } }}// Function to insert edges in the graphvoid insertEdges(int u, int v, vector<int> adj[]){ adj[u].push_back(v); adj[v].push_back(u);}// Driver Codeint main(){ int n = 5, m = 5; vector<int> adj[n + 1]; // Insert edges insertEdges(1, 4, adj); insertEdges(3, 4, adj); insertEdges(5, 4, adj); insertEdges(3, 2, adj); insertEdges(1, 5, adj); // Function call printLexoSmall(adj, n); return 0;} |
Java
// Java program to print the lexcicographically // smallest path starting from 1 import java.util.*;public class GFG{ // Function to print the smallest lexicographically // BFS path starting from 1 static void printLexoSmall(Vector<Vector<Integer>> adj, int n) { // Visited array boolean[] vis = new boolean[n + 1]; // Minimum Heap Vector<Integer> Q = new Vector<Integer>(); // First one visited vis[1] = true; Q.add(1); // Iterate till all nodes are visited while (Q.size() > 0) { // Get the top element int now = Q.get(0); // Pop the element Q.remove(0); // Print the current node System.out.print(now + " "); // Find adjacent nodes for(int p : adj.get(now)) { // If not visited if (!vis[p]) { // Push Q.add(p); Collections.sort(Q); // Mark as visited vis[p] = true; } } } } // Function to insert edges in the graph static void insertEdges(int u, int v, Vector<Vector<Integer>> adj) { adj.get(u).add(v); adj.get(v).add(u); } // Driver code public static void main(String[] args) { int n = 5; Vector<Vector<Integer>> adj = new Vector<Vector<Integer>>(); for(int i = 0; i < n + 1; i++) { adj.add(new Vector<Integer>()); } // Insert edges insertEdges(1, 4, adj); insertEdges(3, 4, adj); insertEdges(5, 4, adj); insertEdges(3, 2, adj); insertEdges(1, 5, adj); // Function call printLexoSmall(adj, n); }}// This code is contributed by divyeshrabadiya07. |
Python3
# Python program to print the lexcicographically # smallest path starting from 1 # Function to print the smallest lexicographically # BFS path starting from 1 def printLexoSmall(adj, n): # Visited array vis = [False for i in range(n + 1)] # Minimum Heap Q = [] # First one visited vis[1] = True; Q.append(1) # Iterate till all nodes are visited while(len(Q) != 0): # Get the top element now = Q[0] # Pop the element Q.pop(0) # Print the current node print(now, end = " ") # Find adjacent nodes for p in adj[now]: # If not visited if(not vis[p]): # Push Q.append(p) Q.sort() # Mark as visited vis[p] = True# Function to insert edges in the graph def insertEdges(u, v, adj): adj[u].append(v) adj[v].append(u)# Driver coden = 5m = 5adj = [[] for i in range(n + 1)]# Insert edges insertEdges(1, 4, adj)insertEdges(3, 4, adj)insertEdges(5, 4, adj)insertEdges(3, 2, adj)insertEdges(1, 5, adj)# Function call printLexoSmall(adj, n)# This code is contributed by avanitrachhadiya2155 |
C#
// C# program to print the lexcicographically // smallest path starting from 1 using System;using System.Collections.Generic;class GFG { // Function to print the smallest lexicographically // BFS path starting from 1 static void printLexoSmall(List<List<int>> adj, int n) { // Visited array bool[] vis = new bool[n + 1]; // Minimum Heap List<int> Q = new List<int>(); // First one visited vis[1] = true; Q.Add(1); // Iterate till all nodes are visited while (Q.Count > 0) { // Get the top element int now = Q[0]; // Pop the element Q.RemoveAt(0); // Print the current node Console.Write(now + " "); // Find adjacent nodes foreach (int p in adj[now]) { // If not visited if (!vis[p]) { // Push Q.Add(p); Q.Sort(); // Mark as visited vis[p] = true; } } } } // Function to insert edges in the graph static void insertEdges(int u, int v, List<List<int>> adj) { adj[u].Add(v); adj[v].Add(u); } // Driver code static void Main() { int n = 5; List<List<int>> adj = new List<List<int>>(); for(int i = 0; i < n + 1; i++) { adj.Add(new List<int>()); } // Insert edges insertEdges(1, 4, adj); insertEdges(3, 4, adj); insertEdges(5, 4, adj); insertEdges(3, 2, adj); insertEdges(1, 5, adj); // Function call printLexoSmall(adj, n); }}// This code is contributed by divyesh072019. |
Javascript
<script>// JavaScript program to print the lexcicographically// smallest path starting from 1// Function to print the smallest lexicographically // BFS path starting from 1function printLexoSmall(adj,n){ // Visited array let vis = new Array(n + 1); for(let i=0;i<n+1;i++) { vis[i]=false; } // Minimum Heap let Q = []; // First one visited vis[1] = true; Q.push(1); // Iterate till all nodes are visited while (Q.length > 0) { // Get the top element let now = Q[0]; // Pop the element Q.shift(); // Print the current node document.write(now + " "); // Find adjacent nodes for(let p=0;p< adj[now].length;p++) { // If not visited if (!vis[adj[now][p]]) { // Push Q.push(adj[now][p]); Q.sort(function(a,b){return a-b;}); // Mark as visited vis[adj[now][p]] = true; } } }}// Function to insert edges in the graphfunction insertEdges(u,v,adj){ adj[u].push(v); adj[v].push(u);}n = 5;let adj = [];for(let i = 0; i < n + 1; i++){ adj.push([]);}// Insert edgesinsertEdges(1, 4, adj);insertEdges(3, 4, adj);insertEdges(5, 4, adj);insertEdges(3, 2, adj);insertEdges(1, 5, adj);// Function callprintLexoSmall(adj, n);// This code is contributed by ab2127</script> |
Output:
1 4 3 2 5
Time Complexity: O(N*logN), as we are using a loop to traverse N times and in each traversal, we are doing a priority queue operation which will cost logN time. Where N is the total number of nodes in the tree.
Auxiliary Space: O(N), as we are using extra space for vis and priority queue.
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