Find K vertices in the graph which are connected to at least one of remaining vertices

Given a connected graph with N vertices. The task is to select k(k must be less than or equals to n/2, not necessarily minimum) vertices from the graph such that all these selected vertices are connected to at least one of the non selected vertex. In case of multiple answers print any one of them.

Examples:

Input :

Output : 1
Vertex 1 is connected to all other non selected vertices. Here
{1, 2}, {2, 3}, {3, 4}, {1, 3}, {1, 4}, {2, 4} are also the valid answers



Input :

Output : 1 3
Vertex 1, 3 are connected to all other non selected vertices. {2, 4} is also a valid answer.

Efficient Approach: An efficient way is to find vertices which are even level and odd level using simple dfs or bfs function. Then if the verices at odd level are less than the vertices at even level then print odd level vertices. Otherwise, print even level vertices.

Below is the implementation of the above approach:

C++

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// C++ program to find K vertices in 
// the graph which are connected to at 
// least one of remaining vertices
#include <bits/stdc++.h>
using namespace std;
#define N 200005
  
// To store graph
int n, m, vis[N];
vector<int> gr[N];
vector<int> v[2];
  
// Function to add edge
void add_edges(int x, int y)
{
    gr[x].push_back(y);
    gr[y].push_back(x);
}
  
// Function to find level of each node
void dfs(int x, int state)
{
    // Push the vertex in respected level
    v[state].push_back(x);
  
    // Make vertex visited
    vis[x] = 1;
  
    // Traverse for all it's child nodes
    for (auto i : gr[x])
        if (vis[i] == 0)
            dfs(i, state ^ 1);
}
  
// Function to print vertices
void Print_vertices()
{
    // If odd level vertices are less
    if (v[0].size() < v[1].size()) {
        for (auto i : v[0])
            cout << i << " ";
    }
    // If even level vertices are less
    else {
        for (auto i : v[1])
            cout << i << " ";
    }
}
  
// Driver code
int main()
{
    int n = 4, m = 3;
  
    // Add edges
    add_edges(1, 2);
    add_edges(2, 3);
    add_edges(3, 4);
  
    // Function call
    dfs(1, 0);
  
    Print_vertices();
  
    return 0;
}

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Python3

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# Python3 program to find K vertices in
# the graph which are connected to at
# least one of remaining vertices
  
N = 200005
  
# To store graph
n, m, =0,0
vis=[0 for i in range(N)]
gr=[[] for i in range(N)]
v=[[] for i in range(2)]
  
# Function to add edge
def add_edges(x, y):
    gr[x].append(y)
    gr[y].append(x)
  
# Function to find level of each node
def dfs(x, state):
  
    # Push the vertex in respected level
    v[state].append(x)
  
    # Make vertex visited
    vis[x] = 1
  
    # Traverse for all it's child nodes
    for i in gr[x]:
        if (vis[i] == 0):
            dfs(i, state ^ 1)
  
  
# Function to prvertices
def Print_vertices():
  
    # If odd level vertices are less
    if (len(v[0]) < len(v[1])):
        for i in v[0]:
            print(i,end=" ")
    # If even level vertices are less
    else:
        for i in v[1]:
            print(i,end=" ")
  
# Driver code
  
n = 4
m = 3
  
# Add edges
add_edges(1, 2)
add_edges(2, 3)
add_edges(3, 4)
  
# Function call
dfs(1, 0)
  
Print_vertices()
  
# This code is contributed by mohit kumar 29

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Output:

2 4

Time Complexity : O(V+E)
Where V is the number of vertices and E is the number of edges in the graph.



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Improved By : mohit kumar 29



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