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Maximum number of edges to be removed to contain exactly K connected components in the Graph
  • Last Updated : 24 Dec, 2020

Given an undirected graph G with N nodes, M edges, and an integer K, the task is to find the maximum count of edges that can be removed such that there remains exactly K connected components after the removal of edges. If the graph cannot contain K connect components, print -1.

Examples:

Input: N = 4, M = 3, K = 2, Edges[][] = {{1, 2}, {2, 3}, {3, 4}} 
 

Output: 1
Explanation:
One possible way is to remove edge [1, 2]. Then there will be 2 connect components as shown below: 
 



 

Input: N = 3, M = 3, K = 3, Edges[][] = {{1, 2}, {2, 3}, {3, 1}} 
 

Output: 3
Explanation: All edges can be removed to make 3 connected components as shown below: 
 

 

Approach: To solve the given problem, count the number of connected components present in the given graph. Let the count be C. Observe that if C is greater than K then no possible edge removal can generate K connected components as the number of connected components will only increase. Otherwise, the answer will always exist.



Following observations need to be made in order to solve the problem: 
 

  • Suppose C1, C2, …, Cc, are the number of node in each connected component. Then, each component must have edges as C1 – 1, C2 – 1, …, C-1 after edges are removed. Therefore, 
     

C1 – 1 + C2 – 1 + … + Cc – 1 = C1 + C2 + … + Cc – C = N – C, where N is the number of nodes. 
 

  • The above condition will give us the C connected components by removing M – (N – C) edges as N – C edges are needed to make C components. To get K components, (K – C) more edges must be removed.
  • Hence, the total count of edges to be removed is given by: 
     

M – (N – C) + (K – C) = M – N + K 
 

Follow the steps below to solve the problem:  

  1. Count the number of connected components present in the given graph. Let the count be C.
  2. If C is greater than K, print -1.
  3. Else print M – N + K where N is the number f nodes, M is the number of edges and K is the required number of connected components.

Below is the implementation of the above approach: 

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
class Graph
{
  public:
    int V;
    map<int, vector<int>> adj;
 
    Graph(int);
    void addEdge(int, int);
    void DFS(int, vector<bool> &);
} * g;
 
// Constructor
Graph::Graph(int V)
{
   
  // No. of vertices
  this->V = V;
   
  // Dictionary of lists
  for(int i = 1; i <= V; i++)
    adj[i] = vector<int>();
}
 
// Function to add edge
// in the graph
void Graph::addEdge(int v, int w)
{
  adj[v].push_back(w);
  adj[w].push_back(v);
}
 
// Function to perform DFS
void Graph::DFS(int s, vector<bool> &visited)
{
   
  // Create a stack for DFS
  stack<int> stack;
 
  // Push the current source node
  stack.push(s);
  while (!stack.empty())
  {
     
    // Pop a vertex from stack
    // and print it
    s = stack.top();
    stack.pop();
 
    // Traverse adjacent vertices
    // of the popped vertex s
    for(auto node : adj[s])
    {
      if (!visited[node])
      {
         
        // If adjacent is unvisited,
        // push it to the stack
        visited[node] = true;
        stack.push(node);
      }
    }
  }
}
 
// Function to return the count
// edges removed
void countRemovedEdges(int N, int M, int K)
{
  int C = 0;
 
  // Initially mark all verices
  // as not visited
  vector<bool> visited(g->V + 1, false);
 
  for(int node = 1; node <= N; node++)
  {
     
    // If node is unvisited
    if (!visited[node])
    {
       
      // Increment Connected
      // component count by 1
      C = C + 1;
 
      // Perform DFS Traversal
      g->DFS(node, visited);
 
      // Print the result
      if (C <= K)
        cout << M - N + K << endl;
      else
        cout << -1 << endl;
    }
  }
}
 
// Driver Code
int main(int argc, char const *argv[])
{
  int N = 4, M = 3, K = 2;
 
  // Create Graph
  g = new Graph(N);
 
  // Given Edges
  g->addEdge(1, 2);
  g->addEdge(2, 3);
  g->addEdge(3, 4);
 
  // Function Call
  countRemovedEdges(N, M, K);
}
 
// This code is contributed by sanjeev2552

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Python3

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# Python3 program for the above approach
 
class Graph:
 
    # Constructor
    def __init__(self, V):
 
        # No. of vertices
        self.V = V
 
        # Dictionary of lists
        self.adj = {i: [] for i in range(1, V + 1)}
 
    # Function to add edge
    # in the graph
    def addEdge(self, v, w):
        self.adj[v].append(w)
        self.adj[w].append(v)
 
    # Function to perform DFS
    def DFS(self, s, visited):
 
    # Create a stack for DFS
        stack = []
 
        # Push the current source node
        stack.append(s)
        while (len(stack)):
 
            # Pop a vertex from stack
            # and print it
            s = stack[-1]
            stack.pop()
 
            # Traverse adjacent vertices
            # of the popped vertex s
            for node in self.adj[s]:
                if (not visited[node]):
 
                    # If adjacent is unvisited,
                    # push it to the stack
                    visited[node] = True
                    stack.append(node)
 
# Function to return the count
# edges removed
def countRemovedEdges(N, M, K):
 
    C = 0
 
    # Initially mark all verices
    # as not visited
    visited = [False for i in range(g.V + 1)]
 
    for node in range(1, N + 1):
 
        # If node is unvisited
        if (not visited[node]):
 
            # Increment Connected
            # component count by 1
            C = C + 1
 
            # Perform DFS Traversal
            g.DFS(node, visited)
 
    # Print the result
    if C <= K:
        print(M - N + K)
    else:
        print(-1)
 
 
# Driver Code
 
N, M, K = 4, 3, 2
 
# Create Graph
g = Graph(N)
 
# Given Edges
g.addEdge(1, 2)
g.addEdge(2, 3)
g.addEdge(3, 4)
 
# Function Call
countRemovedEdges(N, M, K)

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Output: 

1

 

Time Complexity: O(N + M)
Auxiliary Space: O(M + N)

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