Given n friends and their friendship relations, find the total number of groups that exist. And the number of ways of new groups that can be formed consisting of people from every existing group.

If no relation is given for any person then that person has no group and singularly forms a group. If a is a friend of b and b is a friend of c, then a b and c form a group.

**Examples:**

Input : Number of people = 6 Relations : 1 - 2, 3 - 4 and 5 - 6 Output: Number of existing Groups = 3 Number of new groups that can be formed = 8 Explanation: The existing groups are (1, 2), (3, 4), (5, 6). The new 8 groups that can be formed by considering a member of every group are (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (2, 3, 5), (2, 3, 6), (2, 4, 5) and (2, 4, 6). Input: Number of people = 6 Relations : 1 - 2 and 2 - 3 Output: Number of existing Groups = 2 Number of new groups that can be formed = 3 Explanation: The existing groups are (1, 2, 3) and (4). The new groups that can be formed by considering a member of every group are (1, 4), (2, 4), (3, 4).

To count number of groups, we need to simply count connected components in the given undirected graph. Counting connected components can be easily done using DFS or BFS.

Since this is an undirected graph, the number of times a Depth First Search starts from an unvisited vertex for every friend is equal to the number of groups formed.

To count number of ways in which we form new groups can be done using simply formula which is (N1)*(N2)*….(Nn) where Ni is the no of people in i-th group.

## C++

`// CPP program to count number of existing ` `// groups and number of new groups that can ` `// be formed. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `class` `Graph { ` ` ` `int` `V; ` `// No. of vertices ` ` ` ` ` `// Pointer to an array containing ` ` ` `// adjacency lists ` ` ` `list<` `int` `>* adj; ` ` ` ` ` `int` `countUtil(` `int` `v, ` `bool` `visited[]); ` `public` `: ` ` ` `Graph(` `int` `V); ` `// Constructor ` ` ` ` ` `// function to add an edge to graph ` ` ` `void` `addRelation(` `int` `v, ` `int` `w); ` ` ` `void` `countGroups(); ` `}; ` ` ` `Graph::Graph(` `int` `V) ` `{ ` ` ` `this` `->V = V; ` ` ` `adj = ` `new` `list<` `int` `>[V]; ` `} ` ` ` `// Adds a relation as a two way edge of ` `// undirected graph. ` `void` `Graph::addRelation(` `int` `v, ` `int` `w) ` `{ ` ` ` `// Since indexing is 0 based, reducing ` ` ` `// edge numbers by 1. ` ` ` `v--; ` ` ` `w--; ` ` ` `adj[v].push_back(w); ` ` ` `adj[w].push_back(v); ` `} ` ` ` `// Returns count of not visited nodes reachable ` `// from v using DFS. ` `int` `Graph::countUtil(` `int` `v, ` `bool` `visited[]) ` `{ ` ` ` `int` `count = 1; ` ` ` `visited[v] = ` `true` `; ` ` ` `for` `(` `auto` `i=adj[v].begin(); i!=adj[v].end(); ++i) ` ` ` `if` `(!visited[*i]) ` ` ` `count = count + countUtil(*i, visited); ` ` ` `return` `count; ` `} ` ` ` `// A DFS based function to Count number of ` `// existing groups and number of new groups ` `// that can be formed using a member of ` `// every group. ` `void` `Graph::countGroups() ` `{ ` ` ` `// Mark all the vertices as not visited ` ` ` `bool` `* visited = ` `new` `bool` `[V]; ` ` ` `memset` `(visited, 0, V*` `sizeof` `(` `int` `)); ` ` ` ` ` `int` `existing_groups = 0, new_groups = 1; ` ` ` `for` `(` `int` `i = 0; i < V; i++) ` ` ` `{ ` ` ` `// If not in any group. ` ` ` `if` `(visited[i] == ` `false` `) ` ` ` `{ ` ` ` `existing_groups++; ` ` ` ` ` `// Number of new groups that ` ` ` `// can be formed. ` ` ` `new_groups = new_groups * ` ` ` `countUtil(i, visited); ` ` ` `} ` ` ` `} ` ` ` ` ` `if` `(existing_groups == 1) ` ` ` `new_groups = 0; ` ` ` ` ` `cout << ` `"No. of existing groups are "` ` ` `<< existing_groups << endl; ` ` ` `cout << ` `"No. of new groups that can be"` ` ` `" formed are "` `<< new_groups ` ` ` `<< endl; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 6; ` ` ` ` ` `// Create a graph given in the above diagram ` ` ` `Graph g(n); ` `// total 6 people ` ` ` `g.addRelation(1, 2); ` `// 1 and 2 are friends ` ` ` `g.addRelation(3, 4); ` `// 3 and 4 are friends ` ` ` `g.addRelation(5, 6); ` `// 5 and 6 are friends ` ` ` ` ` `g.countGroups(); ` ` ` ` ` `return` `0; ` `} ` |

## Python3

# Python3 program to count number of

# existing groups and number of new

# groups that can be formed.

class Graph:

def __init__(self, V):

self.V = V

self.adj = [[] for i in range(V)]

# Adds a relation as a two way

# edge of undirected graph.

def addRelation(self, v, w):

# Since indexing is 0 based,

# reducing edge numbers by 1.

v -= 1

w -= 1

self.adj[v].append(w)

self.adj[w].append(v)

# Returns count of not visited

# nodes reachable from v using DFS.

def countUtil(self, v, visited):

count = 1

visited[v] = True

i = 0

while i != len(self.adj[v]):

if (not visited[self.adj[v][i]]):

count = count + self.countUtil(self.adj[v][i],

visited)

i += 1

return count

# A DFS based function to Count number

# of existing groups and number of new

# groups that can be formed using a

# member of every group.

def countGroups(self):

# Mark all the vertices as

# not visited

visited = [0] * self.V

existing_groups = 0

new_groups = 1

for i in range(self.V):

# If not in any group.

if (visited[i] == False):

existing_groups += 1

# Number of new groups that

# can be formed.

new_groups = (new_groups *

self.countUtil(i, visited))

if (existing_groups == 1):

new_groups = 0

print(“No. of existing groups are”,

existing_groups)

print(“No. of new groups that”,

“can be formed are”, new_groups)

# Driver code

if __name__ == ‘__main__’:

n = 6

# Create a graph given in the above diagram

g = Graph(n) # total 6 people

g.addRelation(1, 2) # 1 and 2 are friends

g.addRelation(3, 4) # 3 and 4 are friends

g.addRelation(5, 6) # 5 and 6 are friends

g.countGroups()

# This code is contributed by PranchalK

**Output:**

No. of existing groups are 3 No. of new groups that can be formed are 8

Time complexity: O(N + R) where N is the the number of people and R is the number of relations.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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