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++
// 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 2void 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 Codeint main(){ int N = 5, K = 3; constGraphWithCon(N, K); return 0;} |
C
// 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 2void 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 Codeint main(){ int N = 5, K = 3; constGraphWithCon(N, K); return 0;} |
Java
// Java program to implement// the above approachimport 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 2static 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 Codepublic static void main(String[] args){ int N = 5, K = 3; constGraphWithCon(N, K);}}// This code is contributed by 29AjayKumar |
Python3
# 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 2def 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 codeif __name__ == '__main__': N = 5 K = 3 constGraphWithCon(N, K)# This code is contributed by MuskanKalra1 |
C#
// C# program to implement// the above approachusing 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 2static 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 Codepublic static void Main(String[] args){ int N = 5, K = 3; constGraphWithCon(N, K);}}// This code is contributed by 29AjayKumar |
Javascript
<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 2function 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 Codevar N = 5, K = 3;constGraphWithCon(N, K);</script> |
Output:
1 2 1 3 1 4 1 5 2 3 2 4 2 5
Time Complexity: O(N2)
Auxiliary Space: O(N2)
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