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# Minimize count of connections required to be rearranged to make all the computers connected

• Difficulty Level : Hard
• Last Updated : 28 Mar, 2023

Given an integer N, denoting the number of computers connected by cables forming a network and a 2D array connections[][], with each row (i, j) representing a connection between ith and jth computer, the task is to connect all the computers either directly or indirectly by removing any of the given connections and connecting two disconnected computers If it’s not possible to connect all the computers, print -1. Otherwise, print the minimum number of such operations required.

Examples:

Input: N = 4, connections[][] = {{0, 1}, {0, 2}, {1, 2}}
Output: 1
Explanation: Remove the connection between computers 1 and 2 and connect the computers 1 and 3.

Input: N = 5, connections[][] = {{0, 1}, {0, 2}, {3, 4}, {2, 3}}
Output: 0

## Using Disjoint Set And Adjacency List :

```      Here We use disjoint set to calculate extra connections between computers then using disjoint set method
call findupar() then we will calculate the total number of components of graph.
Now.
we have k i.e extra edges
&
n i.e number of components in graph.

if((totalcomponents-1)<=k)return totalcomponents-1; // totalcomponents-1 must be less then extra edges in all
components to connect all to each other.
else return -1;     // if not return -1.```

Follow the steps below to solve the problem:

•  make adjacency list from given 2D array.
•  call disjoint class with given number of computers.
•  calculate total number of components in graph using dfs.
•  last step check totalcomponents – 1 <= totalnumberofextraedges.

Below is the implementation of the above approach:

## C++

 `#include ` `#include `   `using` `namespace` `std;`   `class` `DisjointSet {` `    ``vector<``int``> rank, parent, size;`   `public``:` `    ``DisjointSet(``int` `n)` `    ``{` `        ``rank.resize(n + 1, 0);` `        ``parent.resize(n + 1);` `        ``size.resize(n + 1);` `        ``for` `(``int` `i = 0; i <= n; i++) {` `            ``parent[i] = i;` `            ``size[i] = 1;` `        ``}` `    ``}`   `    ``int` `findUPar(``int` `node)` `    ``{` `        ``if` `(node == parent[node])` `            ``return` `node;` `        ``return` `parent[node] = findUPar(parent[node]);` `    ``}`   `    ``void` `unionByRank(``int` `u, ``int` `v)` `    ``{` `        ``int` `ulp_u = findUPar(u);` `        ``int` `ulp_v = findUPar(v);` `        ``if` `(ulp_u == ulp_v)` `            ``return``;` `        ``if` `(rank[ulp_u] < rank[ulp_v]) {` `            ``parent[ulp_u] = ulp_v;` `        ``}` `        ``else` `if` `(rank[ulp_v] < rank[ulp_u]) {` `            ``parent[ulp_v] = ulp_u;` `        ``}` `        ``else` `{` `            ``parent[ulp_v] = ulp_u;` `            ``rank[ulp_u]++;` `        ``}` `    ``}`   `    ``void` `unionBySize(``int` `u, ``int` `v)` `    ``{` `        ``int` `ulp_u = findUPar(u);` `        ``int` `ulp_v = findUPar(v);` `        ``if` `(ulp_u == ulp_v)` `            ``return``;` `        ``if` `(size[ulp_u] < size[ulp_v]) {` `            ``parent[ulp_u] = ulp_v;` `            ``size[ulp_v] += size[ulp_u];` `        ``}` `        ``else` `{` `            ``parent[ulp_v] = ulp_u;` `            ``size[ulp_u] += size[ulp_v];` `        ``}` `    ``}` `};`   `void` `dfs(``int` `start, vector<``int``> adj[], vector<``bool``>& vis)` `{` `    ``if` `(vis[start])` `        ``return``;` `    ``vis[start] = 1;` `    ``for` `(``auto` `it : adj[start]) {` `        ``dfs(it, adj, vis);` `    ``}` `}`   `int` `main(){` `    ``int` `n = 6;` `    ``vector > connections{` `      ``{0,1},{0,2},{0,3},{1,2},{1,3}` `    ``};` `    ``vector<``int``> adj[n + 1];` `    ``for` `(``int` `i = 0; i < connections.size(); i++) {` `        ``adj[connections[i][0]].push_back(connections[i][1]);` `        ``adj[connections[i][1]].push_back(connections[i][0]);` `    ``}` `    ``DisjointSet ds(n);` `    ``int` `extra = 0;` `    ``for` `(``int` `i = 0; i < connections.size(); i++) {` `        ``if` `(ds.findUPar(connections[i][0])` `            ``== ds.findUPar(connections[i][1]))` `            ``++extra;` `        ``ds.unionBySize(connections[i][0],` `                       ``connections[i][1]);` `    ``}` `    ``vector<``bool``> vis(n, 0);` `    ``dfs(0, adj, vis);` `    ``int` `cnt = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(!vis[i]) {` `            ``++cnt;` `            ``dfs(i, adj, vis);` `        ``}` `    ``}` `    ``if` `(cnt <= extra) {` `        ``cout << cnt << endl;` `    ``}` `    ``else` `{` `        ``cout << ``"All computer can not be connect "` `             ``<< endl;` `        ``cout<<-1<

Output

`2`

Complexity Analysis :

TimeComplexity : O(n) i.e DFS . the disjoint set function work in almost constant time O(4alpha).

Space Complexity : O (V + E).

## Minimize count of connections using Minimum Spanning Tree

The idea is to use a concept similar to that of Minimum Spanning Tree, as in a Graph with N nodes, only N – 1 edges are required to make all the nodes connected.

Redundant edges = Total edges – [(Number of Nodes – 1) – (Number of components – 1)]

If Redundant edges > (Number of components – 1) : It is clear that there are enough wires that can be used to connect disconnected computers.Otherwise, print -1.

Follow the steps below to solve the problem:

• Initialize an unordered map, say adj to store the adjacency list from the given information about edges.
• Initialize a vector of boolean datatype, say visited, to store whether a node is visited or not.
• Generate the adjacency list and also calculate the number of edges.
• Initialize a variable, say components, to store the count of connected components.
• Traverse the nodes of the graph using DFS to count the number of connected components and store it in the variable components.
• Initialize a variable, say redundant, and store the number of redundant edges using the above formula.
• If redundant edges > components – 1, then the minimum number of required operations is equal to components – 1. Otherwise, print -1.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to visit the nodes of a graph` `void` `DFS(unordered_map<``int``, vector<``int``> >& adj, ``int` `node,` `         ``vector<``bool``>& visited)` `{` `    ``// If current node is already visited` `    ``if` `(visited[node])` `        ``return``;`   `    ``// If current node is not visited` `    ``visited[node] = ``true``;`   `    ``// Recurse for neighbouring nodes` `    ``for` `(``auto` `x : adj[node]) {`   `        ``// If the node is not visited` `        ``if` `(visited[x] == ``false``)` `            ``DFS(adj, x, visited);` `    ``}` `}`   `// Utility function to check if it is` `// possible to connect all computers or not` `int` `makeConnectedUtil(``int` `N, ``int` `connections[][2], ``int` `M)` `{` `    ``// Stores whether a` `    ``// node is visited or not` `    ``vector<``bool``> visited(N, ``false``);`   `    ``// Build the adjacency list` `    ``unordered_map<``int``, vector<``int``> > adj;`   `    ``// Stores count of edges` `    ``int` `edges = 0;`   `    ``// Building adjacency list` `    ``// from the given edges` `    ``for` `(``int` `i = 0; i < M; ++i) {`   `        ``// Add edges` `        ``adj[connections[i][0]].push_back(connections[i][1]);` `        ``adj[connections[i][1]].push_back(connections[i][0]);`   `        ``// Increment count of edges` `        ``edges += 1;` `    ``}`   `    ``// Stores count of components` `    ``int` `components = 0;`   `    ``for` `(``int` `i = 0; i < N; ++i) {`   `        ``// If node is not visited` `        ``if` `(visited[i] == ``false``) {`   `            ``// Increment components` `            ``components += 1;`   `            ``// Perform DFS` `            ``DFS(adj, i, visited);` `        ``}` `    ``}`   `    ``// At least N - 1 edges are required` `    ``if` `(edges < N - 1)` `        ``return` `-1;`   `    ``// Count redundant edges` `    ``int` `redundant = edges - ((N - 1) - (components - 1));`   `    ``// Check if components can be` `    ``// rearranged using redundant edges` `    ``if` `(redundant >= (components - 1))` `        ``return` `components - 1;`   `    ``return` `-1;` `}`   `// Function to check if it is possible` `// to connect all the computers or not` `void` `makeConnected(``int` `N, ``int` `connections[][2], ``int` `M)` `{` `    ``// Stores counmt of minimum` `    ``// operations required` `    ``int` `minOps = makeConnectedUtil(N, connections, M);`   `    ``// Print the minimum number` `    ``// of operations required` `    ``cout << minOps;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given number of computers` `    ``int` `N = 4;`   `    ``// Given set of connections` `    ``int` `connections[][2] = { { 0, 1 }, { 0, 2 }, { 1, 2 } };` `    ``int` `M = ``sizeof``(connections) / ``sizeof``(connections[0]);`   `    ``// Function call to check if it is` `    ``// possible to connect all computers or not` `    ``makeConnected(N, connections, M);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {`   `    ``// Function to visit the nodes of a graph` `    ``public` `static` `void` `    ``DFS(HashMap > adj, ``int` `node,` `        ``boolean` `visited[])` `    ``{` `        ``// If current node is already visited` `        ``if` `(visited[node])` `            ``return``;`   `        ``// If current node is not visited` `        ``visited[node] = ``true``;`   `        ``// Recurse for neighbouring nodes` `        ``for` `(``int` `x : adj.get(node)) {`   `            ``// If the node is not visited` `            ``if` `(visited[x] == ``false``)` `                ``DFS(adj, x, visited);` `        ``}` `    ``}`   `    ``// Utility function to check if it is` `    ``// possible to connect all computers or not` `    ``public` `static` `int` `    ``makeConnectedUtil(``int` `N, ``int` `connections[][], ``int` `M)` `    ``{` `        ``// Stores whether a` `        ``// node is visited or not` `        ``boolean` `visited[] = ``new` `boolean``[N];`   `        ``// Build the adjacency list` `        ``HashMap > adj` `            ``= ``new` `HashMap<>();`   `        ``// Initialize the adjacency list` `        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``adj.put(i, ``new` `ArrayList());` `        ``}`   `        ``// Stores count of edges` `        ``int` `edges = ``0``;`   `        ``// Building adjacency list` `        ``// from the given edges` `        ``for` `(``int` `i = ``0``; i < M; ++i) {`   `            ``// Get neighbours list` `            ``ArrayList l1` `                ``= adj.get(connections[i][``0``]);` `            ``ArrayList l2` `                ``= adj.get(connections[i][``0``]);`   `            ``// Add edges` `            ``l1.add(connections[i][``1``]);` `            ``l2.add(connections[i][``0``]);`   `            ``// Increment count of edges` `            ``edges += ``1``;` `        ``}`   `        ``// Stores count of components` `        ``int` `components = ``0``;`   `        ``for` `(``int` `i = ``0``; i < N; ++i) {`   `            ``// If node is not visited` `            ``if` `(visited[i] == ``false``) {`   `                ``// Increment components` `                ``components += ``1``;`   `                ``// Perform DFS` `                ``DFS(adj, i, visited);` `            ``}` `        ``}`   `        ``// At least N - 1 edges are required` `        ``if` `(edges < N - ``1``)` `            ``return` `-``1``;`   `        ``// Count redundant edges` `        ``int` `redundant` `            ``= edges - ((N - ``1``) - (components - ``1``));`   `        ``// Check if components can be` `        ``// rearranged using redundant edges` `        ``if` `(redundant >= (components - ``1``))` `            ``return` `components - ``1``;`   `        ``return` `-``1``;` `    ``}`   `    ``// Function to check if it is possible` `    ``// to connect all the computers or not` `    ``public` `static` `void` `    ``makeConnected(``int` `N, ``int` `connections[][], ``int` `M)` `    ``{` `        ``// Stores counmt of minimum` `        ``// operations required` `        ``int` `minOps = makeConnectedUtil(N, connections, M);`   `        ``// Print the minimum number` `        ``// of operations required` `        ``System.out.println(minOps);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``// Given number of computers` `        ``int` `N = ``4``;`   `        ``// Given set of connections` `        ``int` `connections[][]` `            ``= { { ``0``, ``1` `}, { ``0``, ``2` `}, { ``1``, ``2` `} };` `        ``int` `M = connections.length;`   `        ``// Function call to check if it is` `        ``// possible to connect all computers or not` `        ``makeConnected(N, connections, M);` `    ``}` `}`   `// This code is contributed by kingash.`

## Python3

 `# Python3 code for the above approach`   `# Function to visit the nodes of a graph`     `def` `DFS(adj, node, visited):`   `    ``# If current node is already visited` `    ``if` `(visited[node]):` `        ``return`   `    ``# If current node is not visited` `    ``visited[node] ``=` `True`   `    ``# Recurse for neighbouring nodes` `    ``if``(node ``in` `adj):` `        ``for` `x ``in` `adj[node]:`   `            ``# If the node is not visited` `            ``if` `(visited[x] ``=``=` `False``):` `                ``DFS(adj, x, visited)`   `# Utility function to check if it is` `# possible to connect all computers or not`     `def` `makeConnectedUtil(N, connections, M):`   `    ``# Stores whether a` `    ``# node is visited or not` `    ``visited ``=` `[``False` `for` `i ``in` `range``(N)]`   `    ``# Build the adjacency list` `    ``adj ``=` `{}`   `    ``# Stores count of edges` `    ``edges ``=` `0`   `    ``# Building adjacency list` `    ``# from the given edges` `    ``for` `i ``in` `range``(M):`   `        ``# Add edges` `        ``if` `(connections[i][``0``] ``in` `adj):` `            ``adj[connections[i][``0``]].append(` `                ``connections[i][``1``])` `        ``else``:` `            ``adj[connections[i][``0``]] ``=` `[]` `        ``if` `(connections[i][``1``] ``in` `adj):` `            ``adj[connections[i][``1``]].append(` `                ``connections[i][``0``])` `        ``else``:` `            ``adj[connections[i][``1``]] ``=` `[]`   `        ``# Increment count of edges` `        ``edges ``+``=` `1`   `    ``# Stores count of components` `    ``components ``=` `0`   `    ``for` `i ``in` `range``(N):`   `        ``# If node is not visited` `        ``if` `(visited[i] ``=``=` `False``):`   `            ``# Increment components` `            ``components ``+``=` `1`   `            ``# Perform DFS` `            ``DFS(adj, i, visited)`   `    ``# At least N - 1 edges are required` `    ``if` `(edges < N ``-` `1``):` `        ``return` `-``1`   `    ``# Count redundant edges` `    ``redundant ``=` `edges ``-` `((N ``-` `1``) ``-` `(components ``-` `1``))`   `    ``# Check if components can be` `    ``# rearranged using redundant edges` `    ``if` `(redundant >``=` `(components ``-` `1``)):` `        ``return` `components ``-` `1`   `    ``return` `-``1`   `# Function to check if it is possible` `# to connect all the computers or not`     `def` `makeConnected(N, connections, M):`   `    ``# Stores counmt of minimum` `    ``# operations required` `    ``minOps ``=` `makeConnectedUtil(N, connections, M)`   `    ``# Print the minimum number` `    ``# of operations required` `    ``print``(minOps)`     `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:`   `    ``# Given number of computers` `    ``N ``=` `4`   `    ``# Given set of connections` `    ``connections ``=` `[[``0``, ``1``], [``0``, ``2``], [``1``, ``2``]]` `    ``M ``=` `len``(connections)`   `    ``# Function call to check if it is` `    ``# possible to connect all computers or not` `    ``makeConnected(N, connections, M)`   `# This code is contributed by ipg2016107`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG {`   `    ``// Function to visit the nodes of a graph` `    ``public` `static` `void` `DFS(Dictionary<``int``, List<``int``> > adj,` `                           ``int` `node, ``bool``[] visited)` `    ``{` `        ``// If current node is already visited` `        ``if` `(visited[node])` `            ``return``;`   `        ``// If current node is not visited` `        ``visited[node] = ``true``;`   `        ``// Recurse for neighbouring nodes` `        ``foreach``(``int` `x ``in` `adj[node])` `        ``{`   `            ``// If the node is not visited` `            ``if` `(visited[x] == ``false``)` `                ``DFS(adj, x, visited);` `        ``}` `    ``}`   `    ``// Utility function to check if it is` `    ``// possible to connect all computers or not` `    ``public` `static` `int` `    ``makeConnectedUtil(``int` `N, ``int``[, ] connections, ``int` `M)` `    ``{` `        ``// Stores whether a` `        ``// node is visited or not` `        ``bool``[] visited = ``new` `bool``[N];`   `        ``// Build the adjacency list` `        ``Dictionary<``int``, List<``int``> > adj` `            ``= ``new` `Dictionary<``int``, List<``int``> >();`   `        ``// Initialize the adjacency list` `        ``for` `(``int` `i = 0; i < N; i++) {` `            ``adj[i] = ``new` `List<``int``>();` `        ``}`   `        ``// Stores count of edges` `        ``int` `edges = 0;`   `        ``// Building adjacency list` `        ``// from the given edges` `        ``for` `(``int` `i = 0; i < M; ++i) {`   `            ``// Get neighbours list` `            ``List<``int``> l1 = adj[connections[i, 0]];` `            ``List<``int``> l2 = adj[connections[i, 0]];`   `            ``// Add edges` `            ``l1.Add(connections[i, 1]);` `            ``l2.Add(connections[i, 0]);`   `            ``// Increment count of edges` `            ``edges += 1;` `        ``}`   `        ``// Stores count of components` `        ``int` `components = 0;`   `        ``for` `(``int` `i = 0; i < N; ++i) {`   `            ``// If node is not visited` `            ``if` `(visited[i] == ``false``) {`   `                ``// Increment components` `                ``components += 1;`   `                ``// Perform DFS` `                ``DFS(adj, i, visited);` `            ``}` `        ``}`   `        ``// At least N - 1 edges are required` `        ``if` `(edges < N - 1)` `            ``return` `-1;`   `        ``// Count redundant edges` `        ``int` `redundant` `            ``= edges - ((N - 1) - (components - 1));`   `        ``// Check if components can be` `        ``// rearranged using redundant edges` `        ``if` `(redundant >= (components - 1))` `            ``return` `components - 1;`   `        ``return` `-1;` `    ``}`   `    ``// Function to check if it is possible` `    ``// to connect all the computers or not` `    ``public` `static` `void` `    ``makeConnected(``int` `N, ``int``[, ] connections, ``int` `M)` `    ``{` `        ``// Stores counmt of minimum` `        ``// operations required` `        ``int` `minOps = makeConnectedUtil(N, connections, M);`   `        ``// Print the minimum number` `        ``// of operations required` `        ``Console.WriteLine(minOps);` `    ``}`   `    ``static` `void` `Main()` `    ``{` `        ``// Given number of computers` `        ``int` `N = 4;`   `        ``// Given set of connections` `        ``int``[, ] connections` `            ``= { { 0, 1 }, { 0, 2 }, { 1, 2 } };` `        ``int` `M = connections.GetLength(0);`   `        ``// Function call to check if it is` `        ``// possible to connect all computers or not` `        ``makeConnected(N, connections, M);` `    ``}` `}`   `// This code is contributed by decode2207.`

## Javascript

 ``

Output

`1`

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

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