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Minimum and maximum number of connected components

Given array A[] of size N representing graph with N nodes. There is an undirected edge between i and A[i] for 1 <= i <= N. The Task for this problem is to find the minimum and maximum number of connected components in the given graph if it is allowed to add an edge between any two nodes from N nodes as long as each node has a degree at most 2 in the graph.

Examples:

Input: A[] = {2, 1, 4, 3, 6, 5}

Output: 1 3

Explanation: maximum number of connected components will be 3. minimum number of connected components will be 1 since we can add edge between node pairs (2, 3) and (4, 5).

screenshot1

Input: A[] = {2, 3, 1, 5, 6, 4}

Output: 2 2

Explanation: maximum number of connected components will be 2 and minimum will be 2 as well since it is not possible to add edge between any two nodes.

screenshot2

Approach: To solve the problem follow the below observations:

Observations:

Depth-First-Search can be used to solve this problem by finding connected components.

  • If any connected component is forming loop then every node of that connected component will have degree 2 (since maximum degree node can have is 2) so it is not possible to add edge between this connected component to any other connected component.
  • Connected component which does not have loop will have nodes with degree less than 2 so we can add edge here with another connected component which is not forming loop. One more thing to observe is that all these connected components which does not have loop can be connected together forming single connected component thus minimizing number of connected components.
  • Maximum number of connected components will not have any edge added

keeping counter for connected components with loop and without loop (let say X and Y) will be enough and the answer for maximum number of connected components will be X + Y and minimum number of connected components will be X + 1 if Y != 0 and X if Y == 0.

Below are the steps for the above approach:

  • Create adjacency list for graph as adj[N + 1] and fill the adjacency table with N edges.
  • Create two counters for counting with and without looped connected components as loopConnectedComponents = 0, withoutLoopConnectedComponents = 0.
  • Create visited vis[N + 1] array for depth-first-search.
  • Create flag = 0 which tells if current connected component during dfs call has loop or not.
  • Iterate over all N nodes and call dfs function for node which is not visited yet.
    • In dfs(V, parent) function mark node V visited by making vis[V] = 1
    • Iterate over neighbours of V as U if it is not visited and it is not the parent node of V then call dfs function for U
    • else if U is not equal to parent and parent is not -1 then mark flag as 1.
    • if flag is 1 then increment in loopConnectedComponents by 1 else increment withoutLoopConnectedComponents by 1.
  • Create variable as maxConnectedComponents = loopConnectedComponents + withoutLoopConnectedComponents.
  • Create variable as minConnectedComponents = loopConnectedComponents and increament it by 1 if withoutLoopConnectedComponents is non zero (since we will unify all connected components without loop into one).
  • print minConnectedComponents and maxConnectedComponents.

Below is the implementation of the above approach:

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
 
// Depth first search
bool dfs(int V, int parent, int& flag, vector<int>& vis,
         vector<vector<int> >& adj)
{
    // Mark visited
    vis[V] = 1;
 
    // Iterating over V's neibours
    for (auto& U : adj[V]) {
 
        // If U is not visited
        if (!vis[U]) {
 
            // Call dfs function for node U
            // with parent V
            dfs(U, V, flag, vis, adj);
        }
 
        // If U is already visited and it is
        // not the last node that we visited
        // then there is loop in connected
        // compoents so mark flag as 1
        else if (U != parent and parent != -1) {
            flag = 1;
        }
    }
}
 
// Function to find minimum and maximum
// connected components
void findMinMaxCC(int A[], int N)
{
 
    // Creating adjacency list
    vector<vector<int> > adj(N + 1);
 
    // Filling adjacency list
    for (int i = 0; i < N; i++) {
 
        // There is undirected edge
        // from i to A[i]
        adj[i + 1].push_back(A[i]);
        adj[A[i]].push_back(i + 1);
    }
 
    // Counters for looped and without looped
    // connected components
    int loopConnectedComponents = 0,
        withoutLoopConnectedComponents = 0;
 
    // Creating visited array for dfs
    vector<int> vis(N + 1, 0);
 
    // flag to check if given connected
    // component has loop
    int flag = 0;
 
    // Iterating for all N nodes
    for (int i = 1; i <= N; i++) {
 
        // If node i is not visited call dfs
        // for that node
        if (!vis[i]) {
 
            // Call dfs function
            dfs(i, -1, flag, vis, adj);
 
            // If flag is true increment looped
            // Connected component counter by 1
            if (flag)
                loopConnectedComponents++;
 
            // If not increment without looped
            // connected components counter by 1
            else
                withoutLoopConnectedComponents++;
 
            // Resetting the flag
            flag = 0;
        }
    }
 
    // maximum connected components
    int maxConnectedComponents
        = loopConnectedComponents
          + withoutLoopConnectedComponents;
 
    // Minimum connected components
    int minConnectedComponents = loopConnectedComponents;
 
    // Unify all connected components which
    // does not have loop as 1 connected component
    if (withoutLoopConnectedComponents != 0)
        minConnectedComponents++;
 
    // Printing answer of minimum and maximum
    // connected components
    cout << minConnectedComponents << " "
         << maxConnectedComponents << endl;
}
 
// Driver Code
int32_t main()
{
 
    // Input 1
    int A[] = { 2, 1, 4, 3, 6, 5 }, N = 6;
 
    // Function Call
    findMinMaxCC(A, N);
 
    // Input 2
    int A1[] = { 2, 3, 1, 5, 6, 4 }, N1 = 6;
 
    // Function Call
    findMinMaxCC(A1, N1);
 
    return 0;
}

Java

import java.util.ArrayList;
import java.util.Arrays;
 
public class ConnectedComponents {
     
    // Depth first search
    static void dfs(int V, int parent, int[] flag, int[] vis, ArrayList<ArrayList<Integer>> adj) {
        // Mark visited
        vis[V] = 1;
 
        // Iterating over V's neighbors
        for (int U : adj.get(V)) {
            // If U is not visited
            if (vis[U] == 0) {
                // Call dfs function for node U with parent V
                dfs(U, V, flag, vis, adj);
            }
            // If U is already visited and it is not the last node that we visited
            // then there is a loop in connected components so mark flag as 1
            else if (U != parent && parent != -1) {
                flag[0] = 1;
            }
        }
    }
 
    // Function to find minimum and maximum connected components
    static void findMinMaxCC(int[] A, int N) {
        // Creating adjacency list
        ArrayList<ArrayList<Integer>> adj = new ArrayList<>(N + 1);
 
        // Filling adjacency list
        for (int i = 0; i <= N; i++) {
            adj.add(new ArrayList<>());
        }
 
        // Filling adjacency list
        for (int i = 0; i < N; i++) {
            // There is an undirected edge from i to A[i]
            adj.get(i + 1).add(A[i]);
            adj.get(A[i]).add(i + 1);
        }
 
        // Counters for looped and without looped connected components
        int[] loopConnectedComponents = {0};
        int[] withoutLoopConnectedComponents = {0};
 
        // Creating visited array for dfs
        int[] vis = new int[N + 1];
 
        // flag to check if the given connected component has a loop
        int[] flag = {0};
 
        // Iterating for all N nodes
        for (int i = 1; i <= N; i++) {
            // If node i is not visited call dfs for that node
            if (vis[i] == 0) {
                // Call dfs function
                dfs(i, -1, flag, vis, adj);
 
                // If flag is true increment looped Connected component counter by 1
                if (flag[0] == 1) {
                    loopConnectedComponents[0]++;
                }
                // If not increment without looped connected components counter by 1
                else {
                    withoutLoopConnectedComponents[0]++;
                }
 
                // Resetting the flag
                flag[0] = 0;
            }
        }
 
        // Maximum connected components
        int maxConnectedComponents = loopConnectedComponents[0] + withoutLoopConnectedComponents[0];
 
        // Minimum connected components
        int minConnectedComponents = loopConnectedComponents[0];
 
        // Unify all connected components which do not have a loop as 1 connected component
        if (withoutLoopConnectedComponents[0] != 0) {
            minConnectedComponents++;
        }
 
        // Printing the answer of minimum and maximum connected components
        System.out.println(minConnectedComponents + " " + maxConnectedComponents);
    }
 
    // Driver Code
    public static void main(String[] args) {
        // Input 1
        int[] A = { 2, 1, 4, 3, 6, 5 };
        int N = 6;
 
        // Function Call
        findMinMaxCC(A, N);
 
        // Input 2
        int[] A1 = { 2, 3, 1, 5, 6, 4 };
        int N1 = 6;
 
        // Function Call
        findMinMaxCC(A1, N1);
    }
}

Python

def dfs(V, parent, flag, vis, adj):
    # Mark visited
    vis[V] = 1
 
    # Iterating over V's neighbors
    for U in adj[V]:
        # If U is not visited
        if not vis[U]:
            # Call dfs function for node U
            # with parent V
            dfs(U, V, flag, vis, adj)
        # If U is already visited and it is
        # not the last node that we visited
        # then there is a loop in connected
        # components so mark flag as 1
        elif U != parent and parent != -1:
            flag[0] = 1
 
 
def findMinMaxCC(A, N):
    # Creating adjacency list
    adj = [[] for _ in range(N + 1)]
 
    # Filling adjacency list
    for i in range(N):
        # There is an undirected edge
        # from i to A[i]
        adj[i + 1].append(A[i])
        adj[A[i]].append(i + 1)
 
    # Counters for looped and without looped
    # connected components
    loopConnectedComponents = 0
    withoutLoopConnectedComponents = 0
 
    # Creating visited array for dfs
    vis = [0] * (N + 1)
 
    # Flag to check if the given connected
    # component has a loop
    flag = [0]
 
    # Iterating for all N nodes
    for i in range(1, N + 1):
        # If node i is not visited, call dfs
        # for that node
        if not vis[i]:
            # Call dfs function
            dfs(i, -1, flag, vis, adj)
 
            # If flag is true, increment looped
            # Connected component counter by 1
            if flag[0]:
                loopConnectedComponents += 1
            # If not, increment without looped
            # connected components counter by 1
            else:
                withoutLoopConnectedComponents += 1
 
            # Resetting the flag
            flag[0] = 0
 
    # Maximum connected components
    maxConnectedComponents = (
        loopConnectedComponents + withoutLoopConnectedComponents
    )
 
    # Minimum connected components
    minConnectedComponents = loopConnectedComponents
 
    # Unify all connected components which
    # do not have a loop as 1 connected component
    if withoutLoopConnectedComponents != 0:
        minConnectedComponents += 1
 
    # Printing answer of minimum and maximum
    # connected components
    print(minConnectedComponents, maxConnectedComponents)
 
 
# Driver Code
if __name__ == "__main__":
    # Input 1
    A = [2, 1, 4, 3, 6, 5]
    N = 6
 
    # Function Call
    findMinMaxCC(A, N)
 
    # Input 2
    A1 = [2, 3, 1, 5, 6, 4]
    N1 = 6
 
    # Function Call
    findMinMaxCC(A1, N1)

C#

using System;
using System.Collections.Generic;
 
public class GFG {
    // Depth first search
    static void DFS(int V, int parent, ref int flag,
                    List<int> vis, List<List<int> > adj)
    {
        // Mark visited
        vis[V] = 1;
 
        // Iterating over V's neighbors
        foreach(int U in adj[V])
        {
            // If U is not visited
            if (vis[U] == 0) {
                // Call DFS function for node U with parent
                // V
                DFS(U, V, ref flag, vis, adj);
            }
            // If U is already visited and it is
            // not the last node that we visited,
            // then there is a loop in connected
            // components, so mark flag as 1
            else if (U != parent && parent != -1) {
                flag = 1;
            }
        }
    }
 
    // Function to find minimum and maximum
    // connected components
    static void FindMinMaxCC(int[] A, int N)
    {
        // Creating adjacency list
        List<List<int> > adj = new List<List<int> >();
        for (int i = 0; i <= N; i++) {
            adj.Add(new List<int>());
        }
 
        // Filling the adjacency list
        for (int i = 0; i < N; i++) {
            // There is an undirected edge from i+1 to A[i]
            adj[i + 1].Add(A[i]);
            adj[A[i]].Add(i + 1);
        }
 
        // Counters for looped and without
        // looped connected components
        int loopConnectedComponents = 0;
        int withoutLoopConnectedComponents = 0;
 
        // Creating a visited array for DFS
        List<int> vis = new List<int>();
        for (int i = 0; i <= N; i++) {
            vis.Add(0);
        }
 
        // Flag to check if a given connected
        // component has a loop
        int flag = 0;
 
        // Iterating for all N nodes
        for (int i = 1; i <= N; i++) {
            // If node i is not visited,
            // call DFS for that node
            if (vis[i] == 0) {
                // Call DFS function
                DFS(i, -1, ref flag, vis, adj);
 
                // If the flag is true, increment looped
                // connected component counter by 1
                if (flag == 1) {
                    loopConnectedComponents++;
                }
                // If not, increment without looped
                // connected components counter by 1
                else {
                    withoutLoopConnectedComponents++;
                }
 
                // Reset the flag
                flag = 0;
            }
        }
 
        // Maximum connected components
        int maxConnectedComponents
            = loopConnectedComponents
              + withoutLoopConnectedComponents;
 
        // Minimum connected components
        int minConnectedComponents
            = loopConnectedComponents;
 
        // Unify all connected components which
        // do not have a loop as 1 connected component
        if (withoutLoopConnectedComponents != 0) {
            minConnectedComponents++;
        }
 
        // Printing the answer of minimum and
        // maximum connected components
        Console.WriteLine(minConnectedComponents + " "
                          + maxConnectedComponents);
    }
 
    public static void Main(string[] args)
    {
        // Input 1
        int[] A = { 2, 1, 4, 3, 6, 5 };
        int N = 6;
 
        // Function Call
        FindMinMaxCC(A, N);
 
        // Input 2
        int[] A1 = { 2, 3, 1, 5, 6, 4 };
        int N1 = 6;
 
        // Function Call
        FindMinMaxCC(A1, N1);
    }
}

Javascript

function dfs(V, parent, vis, adj, flag) {
    // Mark the current node as visited
    vis[V] = true;
    // Iterate over V's neighbors
    for (const U of adj[V]) {
        if (!vis[U]) {
            dfs(U, V, vis, adj, flag);
        }
        // If U is already visited and it's not the parent of V
        // there's a loop
        else if (U !== parent && parent !== -1) {
            flag.flag = true;
        }
    }
}
// Function to find minimum and
// maximum connected components
function GFG(A, N) {
    // Creating adjacency list
    const adj = new Array(N + 1).fill(null).map(() => []);
    for (let i = 0; i < N; i++) {
        adj[i + 1].push(A[i]);
        adj[A[i]].push(i + 1);
    }
    // Counters for looped and without
    // looped connected components
    let loopConnectedComponents = 0;
    let withoutLoopConnectedComponents = 0;
    // Creating visited array for DFS
    const vis = new Array(N + 1).fill(false);
    const flag = { flag: false };
    for (let i = 1; i <= N; i++) {
        // If node i is not visited
        // call DFS for that node
        if (!vis[i]) {
            // Call DFS function
            dfs(i, -1, vis, adj, flag);
            // If flag is true, increment looped connected component counter by 1
            if (flag.flag) {
                loopConnectedComponents++;
            }
            else {
                withoutLoopConnectedComponents++;
            }
            // Resetting the flag
            flag.flag = false;
        }
    }
    // Maximum connected components
    const maxConnectedComponents =
        loopConnectedComponents + withoutLoopConnectedComponents;
    // Minimum connected components
    let minConnectedComponents = loopConnectedComponents;
    if (withoutLoopConnectedComponents !== 0) {
        minConnectedComponents++;
    }
    // Printing the answer of minimum and
    // maximum connected components
    console.log(minConnectedComponents, maxConnectedComponents);
}
// Driver Code
(function () {
    // Input 1
    const A = [2, 1, 4, 3, 6, 5];
    const N = 6;
    // Function Call
    GFG(A, N);
    // Input 2
    const A1 = [2, 3, 1, 5, 6, 4];
    const N1 = 6;
    // Function Call
    GFG(A1, N1);
})();

Output

1 3
2 2

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


 


Reffered: https://www.geeksforgeeks.org

DSA

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