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Print the lexicographically smallest BFS of the graph starting from 1

Last Updated : 16 Jun, 2022
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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 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 
// 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 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# 
// 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 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>

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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