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:
// 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 1 void 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 graph void insertEdges( int u, int v, vector< int > adj[])
{ adj[u].push_back(v);
adj[v].push_back(u);
} // Driver Code int 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 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. |
# 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 code n = 5
m = 5
adj = [[] 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# 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. |
<script> // JavaScript program to print the lexcicographically // smallest path starting from 1 // Function to print the smallest lexicographically // BFS path starting from 1
function 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 graph function 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 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 ab2127 </script> |
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.