Given two positive integers N and K, the task is to construct a simple and connected graph consisting of N vertices with the length of each edge as 1 unit, such that the shortest distance between exactly K pairs of vertices is 2. If it is not possible to construct the graph, then print -1. Otherwise, print the edges of the graph.
Examples:
Input: N = 5, K = 3
Output: { { 1, 2 }, { 1, 3}, { 1, 4 }, { 1, 5 }, { 2, 3 }, { 2, 4 }, { 2, 5 } }
Explanation:
The distance between the pairs of vertices { (3, 4), (4, 5), (3, 5) } is 2.
Input: N = 5, K = 8
Output: -1
Approach: Follow the steps below to solve the problem:
- Since the graph is simple and connected, Therefore, the maximum possible count of edges, say Max is ((N – 1) * (N – 2)) / 2.
- If K is greater than Max, then print -1.
- Initialize an array, say edges[], to store the edges of the graph.
- Otherwise, first connect all the vertices with 1 and store it in edges[], then connect all the pairs of vertices (i, j) such that i >= 2 and j > i and store it in edges[].
- Finally, print the first ((N – 1) + Max – K ) elements of edges[] array.
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <iostream> #include <vector> using namespace std;
// Function to construct the simple and // connected graph such that the distance // between exactly K pairs of vertices is 2 void constGraphWithCon( int N, int K)
{ // Stores maximum possible count
// of edges in a graph
int Max = ((N - 1) * (N - 2)) / 2;
// Base Case
if (K > Max) {
cout << -1 << endl;
return ;
}
// Stores edges of a graph
vector<pair< int , int > > ans;
// Connect all vertices of pairs (i, j)
for ( int i = 1; i < N; i++) {
for ( int j = i + 1; j <= N; j++) {
ans.emplace_back(make_pair(i, j));
}
}
// Print first ((N - 1) + Max - K) elements
// of edges[]
for ( int i = 0; i < (N - 1) + Max - K; i++) {
cout << ans[i].first << " "
<< ans[i].second << endl;
}
} // Driver Code int main()
{ int N = 5, K = 3;
constGraphWithCon(N, K);
return 0;
} |
// C program to implement // the above approach #include <stdio.h> // Function to construct the simple and // connected graph such that the distance // between exactly K pairs of vertices is 2 void constGraphWithCon( int N, int K)
{ // Stores maximum possible count
// of edges in a graph
int Max = ((N - 1) * (N - 2)) / 2;
// Base Case
if (K > Max) {
printf ( "-1" );
return ;
}
// Stores count of edges in a graph
int count = 0;
// Connect all vertices of pairs (i, j)
for ( int i = 1; i < N; i++) {
for ( int j = i + 1; j <= N; j++) {
printf ( "%d %d\n" , i, j);
// Update
count++;
if (count == N * (N - 1) / 2 - K)
break ;
}
if (count == N * (N - 1) / 2 - K)
break ;
}
} // Driver Code int main()
{ int N = 5, K = 3;
constGraphWithCon(N, K);
return 0;
} |
// Java program to implement // the above approach import java.util.*;
class GFG{
static class pair
{ int first, second;
public pair( int first, int second)
{
this .first = first;
this .second = second;
}
} // Function to construct the simple and connected // graph such that the distance between // exactly K pairs of vertices is 2 static void constGraphWithCon( int N, int K)
{ // Stores maximum possible count
// of edges in a graph
int Max = ((N - 1 ) * (N - 2 )) / 2 ;
// Base Case
if (K > Max)
{
System.out.print(- 1 + "\n" );
return ;
}
// Stores edges of a graph
Vector<pair> ans = new Vector<>();
// Connect all vertices of pairs (i, j)
for ( int i = 1 ; i < N; i++)
{
for ( int j = i + 1 ; j <= N; j++)
{
ans.add( new pair(i, j));
}
}
// Print first ((N - 1) + Max - K) elements
// of edges[]
for ( int i = 0 ; i < (N - 1 ) + Max - K; i++)
{
System.out.print(ans.get(i).first + " " +
ans.get(i).second + "\n" );
}
} // Driver Code public static void main(String[] args)
{ int N = 5 , K = 3 ;
constGraphWithCon(N, K);
} } // This code is contributed by 29AjayKumar |
# Python3 program to implement # the above approach # Function to construct the simple and # connected graph such that the distance # between exactly K pairs of vertices is 2 def constGraphWithCon(N, K):
# Stores maximum possible count
# of edges in a graph
Max = ((N - 1 ) * (N - 2 )) / / 2
# Base case
if (K > Max ):
print ( - 1 )
return
# Stores edges of a graph
ans = []
# Connect all vertices of pairs (i, j)
for i in range ( 1 , N):
for j in range (i + 1 , N + 1 ):
ans.append([i, j])
# Print first ((N - 1) + Max - K) elements
# of edges[]
for i in range ( 0 , (N - 1 ) + Max - K):
print (ans[i][ 0 ], ans[i][ 1 ], sep = " " )
# Driver code if __name__ = = '__main__' :
N = 5
K = 3
constGraphWithCon(N, K)
# This code is contributed by MuskanKalra1 |
// C# program to implement // the above approach using System;
using System.Collections.Generic;
public class GFG{
class pair
{ public int first, second;
public pair( int first, int second)
{
this .first = first;
this .second = second;
}
} // Function to construct the simple and connected // graph such that the distance between // exactly K pairs of vertices is 2 static void constGraphWithCon( int N, int K)
{ // Stores maximum possible count
// of edges in a graph
int Max = ((N - 1) * (N - 2)) / 2;
// Base Case
if (K > Max)
{
Console.Write(-1 + "\n" );
return ;
}
// Stores edges of a graph
List<pair> ans = new List<pair>();
// Connect all vertices of pairs (i, j)
for ( int i = 1; i < N; i++)
{
for ( int j = i + 1; j <= N; j++)
{
ans.Add( new pair(i, j));
}
}
// Print first ((N - 1) + Max - K) elements
// of edges[]
for ( int i = 0; i < (N - 1) + Max - K; i++)
{
Console.Write(ans[i].first + " " +
ans[i].second + "\n" );
}
} // Driver Code public static void Main(String[] args)
{ int N = 5, K = 3;
constGraphWithCon(N, K);
} } // This code is contributed by 29AjayKumar |
<script> // Javascript program to implement // the above approach class pair { constructor(first, second)
{
this [0] = first;
this [1] = second;
}
} // Function to construct the simple and connected // graph such that the distance between // exactly K pairs of vertices is 2 function constGraphWithCon(N, K)
{ // Stores maximum possible count
// of edges in a graph
var Max = ((N - 1) * (N - 2)) / 2;
// Base Case
if (K > Max)
{
document.write(-1 + "<br>" );
return ;
}
// Stores edges of a graph
var ans = [];
// Connect all vertices of pairs (i, j)
for ( var i = 1; i < N; i++)
{
for ( var j = i + 1; j <= N; j++)
{
ans.push([i, j]);
}
}
// Print first ((N - 1) + Max - K) elements
// of edges[]
for ( var i = 0; i < (N - 1) + Max - K; i++)
{
document.write(ans[i][0] + " " +
ans[i][1] + "<br>" );
}
} // Driver Code var N = 5, K = 3;
constGraphWithCon(N, K); </script> |
1 2 1 3 1 4 1 5 2 3 2 4 2 5
Time Complexity: O(N2)
Auxiliary Space: O(N2)