Minimum number of edges that need to be added to form a triangle

Last Updated : 24 Mar, 2023

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 triangle
int 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 code
int 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 approach
import 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# implementation of the approach
using 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 approach
function 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 code
let n = 3;

// Storing the edges in an array of arrays
let v = [[1, 2], [1, 3]];

console.log(minEdges(v, n));

Output:

Time complexity : O(n^3)

Space complexity : O(n^2)

Analysis of Algorithms

SamratDe

Improve

Article Tags :

Minimum number of edges that need to be added to form a triangle

Similar Reads

Basics & Prerequisites

Data Structures

Algorithms

Advanced

Interview Preparation

Practice Problem

Thank You!

What kind of Experience do you want to share?