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 = 3Â
Output: { { 1, 2 }, { 1, 3}, { 1, 4 }, { 1, 5 }, { 2, 3 }, { 2, 4 }, { 2, 5 } }Â
Explanation:Â
The distance between the pairs of vertices { (3, 4), (4, 5), (3, 5) } is 2.Â
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Input: N = 5, K = 8Â
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 approachÂ
#include <iostream>#include <vector>using namespace std;Â
// 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){Â
    // Stores maximum possible count    // of edges in a graph    int Max = ((N - 1) * (N - 2)) / 2;Â
    // Base Case    if (K > Max) {        cout << -1 << endl;        return;    }Â
    // Stores edges of a graph    vector<pair<int, int> > ans;Â
    // 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));        }    }Â
    // 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;    }}Â
// Driver Codeint main(){Â Â Â Â int N = 5, K = 3;Â Â Â Â constGraphWithCon(N, K);Â
    return 0;} |
C
// C program to implement// the above approachÂ
#include <stdio.h>Â
// 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){Â
    // Stores maximum possible count    // of edges in a graph    int Max = ((N - 1) * (N - 2)) / 2;Â
    // Base Case    if (K > Max) {        printf("-1");        return;    }Â
    // Stores count of edges in a graph    int count = 0;Â
    // Connect all vertices of pairs (i, j)    for (int i = 1; i < N; i++) {Â
        for (int j = i + 1; j <= N; j++) {Â
            printf("%d %d\n", i, j);Â
            // Update            count++;Â
            if (count == N * (N - 1) / 2 - K)                break;        }Â
        if (count == N * (N - 1) / 2 - K)            break;    }}Â
// Driver Codeint main(){Â Â Â Â int N = 5, K = 3;Â Â Â Â constGraphWithCon(N, K);Â
    return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;Â
class GFG{Â Â Â Â Â static class pair{ Â Â Â Â int first, second; Â Â Â Â Â Â Â Â Â public pair(int first, int second)Â Â Â Â Â { Â Â Â Â Â Â Â Â this.first = first; Â Â Â Â Â Â Â Â this.second = second; Â Â Â Â }Â Â Â } Â
// 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;Â
    // Base Case    if (K > Max)    {        System.out.print(-1 + "\n");        return;    }Â
    // Stores edges of a graph    Vector<pair> ans = new Vector<>();Â
    // 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));        }    }Â
    // 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");    }}Â
// Driver Codepublic static void main(String[] args){Â Â Â Â int N = 5, K = 3;Â Â Â Â Â Â Â Â Â constGraphWithCon(N, K);}}Â
// This code is contributed by 29AjayKumar |
Python3
# Python3 program to implement# the above approachÂ
# 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 = []Â
    # 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 = " ")Â
# Driver codeif __name__ == '__main__':Â Â Â Â Â Â Â Â Â N = 5Â Â Â Â K = 3Â Â Â Â Â Â Â Â Â constGraphWithCon(N, K)Â
# This code is contributed by MuskanKalra1 |
C#
// C# program to implement// the above approachusing System;using System.Collections.Generic;Â
public class GFG{Â Â Â Â Â class pair{ Â Â Â Â public int first, second; Â Â Â Â Â Â Â Â Â public pair(int first, int second)Â Â Â Â Â { Â Â Â Â Â Â Â Â this.first = first; Â Â Â Â Â Â Â Â this.second = second; Â Â Â Â }Â Â Â } Â
// 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;Â
    // Base Case    if (K > Max)    {        Console.Write(-1 + "\n");        return;    }Â
    // Stores edges of a graph    List<pair> ans = new List<pair>();Â
    // 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));        }    }Â
    // 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");    }}Â
// Driver Codepublic static void Main(String[] args){Â Â Â Â int N = 5, K = 3;Â Â Â Â constGraphWithCon(N, K);}}Â
// This code is contributed by 29AjayKumar |
Javascript
<script>Â
// Javascript program to implement// the above approach     class pair{     constructor(first, second)    {        this[0] = first;        this[1] = second;    }} Â
// 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;Â
    // Base Case    if (K > Max)    {        document.write(-1 + "<br>");        return;    }Â
    // Stores edges of a graph    var ans = [];Â
    // 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]);        }    }Â
    // 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>");    }}Â
// Driver Codevar N = 5, K = 3;constGraphWithCon(N, K);Â
</script> |
Output:Â
1 2 1 3 1 4 1 5 2 3 2 4 2 5
Â
Time Complexity: O(N2)Â
Auxiliary Space: O(N2)
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