Given an undirected graph with N vertices and N edges. No two edges connect the same pair of vertices. A triangle is a set of three distinct vertices such that each pair of those vertices is connected by an edge i.e. three distinct vertices u, v and w are a triangle if the graph contains the edges (u, v), (v, w) and (v, u). The task is to find the minimum number of edges needed to be added to the given graph such that the graph contains at least one triangle.
Examples:
Input: 1 / \ 2 3 Output: 1 Input: 1 3 / / 2 4 Output: 2
Approach: Initialize ans = 3 which is the maximum count of edges required to form a triangle. Now, for every possible vertex triplet there are four cases:
- Case 1: If there exist nodes i, j and k such that there is an edge from (i, j), (j, k) and (k, i) then the answer is 0.
- Case 2: If there exist nodes i, j and k such that only two pairs of vertices are connected then a single edge is required to form a triangle. So update ans = min(ans, 1).
- Case 3: Otherwise, if there only a single pair of vertices is connected then ans = min(ans, 2).
- Case 4: When there is no edge then ans = min(ans, 3).
Print the ans in the end. Below is the implementation of the above approach.
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the minimum number// of edges that need to be added to// the given graph such that it// contains at least one triangleint minEdges(vector<pair<int, int> > v, int n){ // adj is the adjacency matrix such that // adj[i][j] = 1 when there is an // edge between i and j vector<vector<int> > adj; adj.resize(n + 1); for (int i = 0; i < adj.size(); i++) adj[i].resize(n + 1, 0); // As the graph is undirected // so there will be an edge // between (i, j) and (j, i) for (int i = 0; i < v.size(); i++) { adj[v[i].first][v[i].second] = 1; adj[v[i].second][v[i].first] = 1; } // To store the required // count of edges int edgesNeeded = 3; // For every possible vertex triplet for (int i = 1; i <= n; i++) { for (int j = i + 1; j <= n; j++) { for (int k = j + 1; k <= n; k++) { // If the vertices form a triangle if (adj[i][j] && adj[j][k] && adj[k][i]) return 0; // If no edges are present if (!(adj[i][j] || adj[j][k] || adj[k][i])) edgesNeeded = min(edgesNeeded, 3); else { // If only 1 edge is required if ((adj[i][j] && adj[j][k]) || (adj[j][k] && adj[k][i]) || (adj[k][i] && adj[i][j])) { edgesNeeded = 1; } // Two edges are required else edgesNeeded = min(edgesNeeded, 2); } } } } return edgesNeeded;}// Driver codeint main(){ // Number of nodes int n = 3; // Storing the edges in a vector of pairs vector<pair<int, int> > v = { { 1, 2 }, { 1, 3 } }; cout << minEdges(v, n); return 0;} |
Java
// Java implementation of the approachimport java.io.*;import java.util.*;class Pair<V, E> { V first; E second; Pair(V first, E second) { this.first = first; this.second = second; }}class GFG { // Function to return the minimum number // of edges that need to be added to // the given graph such that it // contains at least one triangle static int minEdges(Vector<Pair<Integer, Integer>> v, int n) { // adj is the adjacency matrix such that // adj[i][j] = 1 when there is an // edge between i and j int[][] adj = new int[n + 1][n + 1]; // As the graph is undirected // so there will be an edge // between (i, j) and (j, i) for (int i = 0; i < v.size(); i++) { adj[v.elementAt(i).first] [v.elementAt(i).second] = 1; adj[v.elementAt(i).second] [v.elementAt(i).first] = 1; } // To store the required // count of edges int edgesNeeded = 0; // For every possible vertex triplet for (int i = 1; i <= n; i++) { for (int j = i + 1; j <= n; j++) { for (int k = j + 1; k <= n; k++) { // If the vertices form a triangle if (adj[i][j] == 1 && adj[j][k] == 1 && adj[k][i] == 1) return 0; // If no edges are present if (!(adj[i][j] == 1 || adj[j][k] == 1 || adj[k][i] == 1)) edgesNeeded = Math.min(edgesNeeded, 3); else { // If only 1 edge is required if ((adj[i][j] == 1 && adj[j][k] == 1) || (adj[j][k] == 1 && adj[k][i] == 1) || (adj[k][i] == 1 && adj[i][j] == 1)) { edgesNeeded = 1; } // Two edges are required else edgesNeeded = Math.min(edgesNeeded, 2); } } } } return edgesNeeded; } // Driver Code public static void main(String[] args) { // Number of nodes int n = 3; // Storing the edges in a vector of pairs Vector<Pair<Integer, Integer>> v = new Vector<>(Arrays.asList(new Pair<>(1, 2), new Pair<>(1, 3))); System.out.println(minEdges(v, n)); }}// This code is contributed by// sanjeev2552 |
Python3
# Python3 implementation of the approach # Function to return the minimum number # of edges that need to be added to # the given graph such that it # contains at least one triangle def minEdges(v, n) : # adj is the adjacency matrix such that # adj[i][j] = 1 when there is an # edge between i and j adj = dict.fromkeys(range(n + 1)); # adj.resize(n + 1); for i in range(n + 1) : adj[i] = [0] * (n + 1); # As the graph is undirected # so there will be an edge # between (i, j) and (j, i) for i in range(len(v)) : adj[v[i][0]][v[i][1]] = 1; adj[v[i][1]][v[i][0]] = 1; # To store the required # count of edges edgesNeeded = 3; # For every possible vertex triplet for i in range(1, n + 1) : for j in range(i + 1, n + 1) : for k in range(j + 1, n + 1) : # If the vertices form a triangle if (adj[i][j] and adj[j][k] and adj[k][i]) : return 0; # If no edges are present if (not (adj[i][j] or adj[j][k] or adj[k][i])) : edgesNeeded = min(edgesNeeded, 3); else : # If only 1 edge is required if ((adj[i][j] and adj[j][k]) or (adj[j][k] and adj[k][i]) or (adj[k][i] and adj[i][j])) : edgesNeeded = 1; # Two edges are required else : edgesNeeded = min(edgesNeeded, 2); return edgesNeeded; # Driver code if __name__ == "__main__" : # Number of nodes n = 3; # Storing the edges in a vector of pairs v = [ [ 1, 2 ], [ 1, 3 ] ]; print(minEdges(v, n)); # This code is contributed by kanugargng |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class Pair { public int first; public int second; public Pair(int first, int second) { this.first = first; this.second = second; }}class GFG { // Function to return the minimum number // of edges that need to be added to // the given graph such that it // contains at least one triangle static int minEdges(List<Pair> v, int n) { // adj is the adjacency matrix such that // adj[i,j] = 1 when there is an // edge between i and j int[,] adj = new int[n + 1, n + 1]; // As the graph is undirected // so there will be an edge // between (i, j) and (j, i) for (int i = 0; i < v.Count; i++) { adj[v[i].first,v[i].second] = 1; adj[v[i].second,v[i].first] = 1; } // To store the required // count of edges int edgesNeeded = 0; // For every possible vertex triplet for (int i = 1; i <= n; i++) { for (int j = i + 1; j <= n; j++) { for (int k = j + 1; k <= n; k++) { // If the vertices form a triangle if (adj[i, j] == 1 && adj[j, k] == 1 && adj[k, i] == 1) return 0; // If no edges are present if (!(adj[i, j] == 1 || adj[j, k] == 1 || adj[k, i] == 1)) edgesNeeded = Math.Min(edgesNeeded, 3); else { // If only 1 edge is required if ((adj[i, j] == 1 && adj[j, k] == 1) || (adj[j, k] == 1 && adj[k, i] == 1) || (adj[k, i] == 1 && adj[i, j] == 1)) { edgesNeeded = 1; } // Two edges are required else edgesNeeded = Math.Min(edgesNeeded, 2); } } } } return edgesNeeded; } // Driver Code public static void Main(String[] args) { // Number of nodes int n = 3; // Storing the edges in a vector of pairs List<Pair> v = new List<Pair>(); v.Add(new Pair(1, 2)); v.Add(new Pair(1, 3)); Console.WriteLine(minEdges(v, n)); }}// This code is contributed by PrinciRaj1992 |
Javascript
// JavaScript implementation of the approachfunction minEdges(v, n) { // adj is the adjacency matrix such that // adj[i][j] = 1 when there is an // edge between i and j let adj = []; for (let i = 0; i <= n; i++) { adj[i] = []; for (let j = 0; j <= n; j++) { adj[i][j] = 0; } } // As the graph is undirected // so there will be an edge // between (i, j) and (j, i) for (let i = 0; i < v.length; i++) { adj[v[i][0]][v[i][1]] = 1; adj[v[i][1]][v[i][0]] = 1; } // To store the required // count of edges let edgesNeeded = 3; // For every possible vertex triplet for (let i = 1; i <= n; i++) { for (let j = i + 1; j <= n; j++) { for (let k = j + 1; k <= n; k++) { // If the vertices form a triangle if (adj[i][j] && adj[j][k] && adj[k][i]) return 0; // If no edges are present if (!(adj[i][j] || adj[j][k] || adj[k][i])) { edgesNeeded = Math.min(edgesNeeded, 3); } else { // If only 1 edge is required if ( (adj[i][j] && adj[j][k]) || (adj[j][k] && adj[k][i]) || (adj[k][i] && adj[i][j]) ) { edgesNeeded = 1; } else { // Two edges are required edgesNeeded = Math.min(edgesNeeded, 2); } } } } } return edgesNeeded;}// Driver codelet n = 3;// Storing the edges in an array of arrayslet v = [[1, 2], [1, 3]];console.log(minEdges(v, n)); |
Output:
1
Time complexity : O(n^3)
Space complexity : O(n^2)
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