Given two integers N and K. The task is to find a connected graph with N vertices such that there are exactly K pairs (i, j) where the shortest distance between them is 2. If no such graph exists then print -1. Note:
- The first-line output should be the number of edges(say m) in the graph and the next m lines should contain two numbers represents the edge between the vertices.
- In case of multiple answers print any of them.
Examples:
Input: N = 5, K = 3 Output: 7 1 2 1 3 1 4 1 5 3 4 3 5 4 5
Input: N = 5, K = 8 Output: -1
Input: N = 5, K = 8 Output: -1Approach: An N-vertices connected graph has at least N-1 edges. The shortest distance of each pair is equal to 1. So obviously, it is clear that there doesn’t exist a solution if K > N * (N – 1) / 2 – (N – 1) = (N – 1) * (N – 2) / 2. Conversely, it can be shown that there exists a solution if K ? (N – 1) * (N – 2) / 2 by constructing a graph that satisfies the condition. First, let’s consider the graph where each vertex is connected with all the other vertices then the shortest between any two vertices is 1. Now remove any K edges then there exist exactly K such pairs. Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to find the required graphvoid connected_graph(int n, int k){ // If no such graph exists if (k > (n - 1) * (n - 2) / 2) { cout << -1 << endl; return; } // Consider edge between all vertices bool isEdge[n][n] = {}; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) isEdge[i][j] = true; } // Remove K vertices int cnt = 0; for (int i = 1; i < n; i++) { for (int j = i + 1; j < n; j++) { if (cnt < k) { isEdge[i][j] = false; cnt++; } } } // Store all the edges vector<pair<int, int> > vec; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { if (isEdge[i][j]) vec.emplace_back(i, j); } } // Print all the edges cout << vec.size() << endl; for (int i = 0; i < vec.size(); i++) { cout << vec[i].first + 1 << " " << vec[i].second + 1 << endl; }}// Driver codeint main(){ int n = 5, k = 3; // Function call connected_graph(n, k); return 0;} |
Java
// Java implementation of the 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 find the required graphstatic void connected_graph(int n, int k){ // If no such graph exists if (k > (n - 1) * (n - 2) / 2) { System.out.println(-1); return; } // Consider edge between all vertices boolean [][]isEdge = new boolean[n][n]; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) isEdge[i][j] = true; } // Remove K vertices int cnt = 0; for (int i = 1; i < n; i++) { for (int j = i + 1; j < n; j++) { if (cnt < k) { isEdge[i][j] = false; cnt++; } } } // Store all the edges Vector<pair> vec = new Vector<>(); for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { if (isEdge[i][j]) vec.add(new pair(i, j)); } } // Print all the edges System.out.println(vec.size()); for (int i = 0; i < vec.size(); i++) { System.out.println(vec.get(i).first + 1 + " " + (vec.get(i).second + 1)); }}// Driver codepublic static void main(String[] args) { int n = 5, k = 3; // Function call connected_graph(n, k);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach import numpy as np;# Function to find the required graph def connected_graph(n, k) : # If no such graph exists if (k > (n - 1) * (n - 2) / 2) : print(-1) ; return; # Consider edge between all vertices isEdge = np.zeros((n, n)); for i in range(n) : for j in range(i + 1, n) : isEdge[i][j] = True; # Remove K vertices cnt = 0; for i in range(1, n) : for j in range(i + 1 , n) : if (cnt < k) : isEdge[i][j] = False; cnt += 1; # Store all the edges vec = []; for i in range(n) : for j in range(i + 1, n) : if (isEdge[i][j]) : vec.append([i, j]); # Print all the edges print(len(vec)); for i in range(len(vec)) : print(vec[i][0] + 1, vec[i][1] + 1); # Driver code if __name__ == "__main__" : n = 5; k = 3; # Function call connected_graph(n, k); # This code is contributed by Ankit Rai |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG{public class pair{ public int first, second; public pair(int first, int second) { this.first = first; this.second = second; } }// Function to find the required graphstatic void connected_graph(int n, int k){ // If no such graph exists if (k > (n - 1) * (n - 2) / 2) { Console.WriteLine(-1); return; } // Consider edge between all vertices bool [,]isEdge = new bool[n, n]; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) isEdge[i, j] = true; } // Remove K vertices int cnt = 0; for (int i = 1; i < n; i++) { for (int j = i + 1; j < n; j++) { if (cnt < k) { isEdge[i, j] = false; cnt++; } } } // Store all the edges List<pair> vec = new List<pair>(); for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { if (isEdge[i, j]) vec.Add(new pair(i, j)); } } // Print all the edges Console.WriteLine(vec.Count); for (int i = 0; i < vec.Count; i++) { Console.WriteLine(vec[i].first + 1 + " " + (vec[i].second + 1)); }}// Driver codepublic static void Main(String[] args) { int n = 5, k = 3; // Function call connected_graph(n, k);}}// This code is contributed by 29AjayKumar |
Javascript
// JS implementationfunction connected_graph(n, k) { // If no such graph exists if (k > (n - 1) * (n - 2) / 2) { console.log(-1); return; } // Consider edge between all vertices let isEdge = new Array(n); for (let i = 0; i < n; i++) { isEdge[i] = new Array(n); for (let j = i + 1; j < n; j++) isEdge[i][j] = true; } // Remove K vertices let cnt = 0; for (let i = 1; i < n; i++) { for (let j = i + 1; j < n; j++) { if (cnt < k) { isEdge[i][j] = false; cnt++; } } } // Store all the edges let vec = []; for (let i = 0; i < n; i++) { for (let j = i + 1; j < n; j++) { if (isEdge[i][j]) vec.push([i, j]); } } // Print all the edges console.log(vec.length); for (let i = 0; i < vec.length; i++) { console.log(vec[i][0] + 1, vec[i][1] + 1); }}// Driver codelet n = 5, k = 3;// Function callconnected_graph(n, k);// This code is contributed by ishankhandelwals. |
Output:
7 1 2 1 3 1 4 1 5 3 4 3 5 4 5
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