# Find the maximum component size after addition of each edge to the graph

Given an array arr[][] which contains the edges of a graph to be used to construct an undirected graph G with N nodes, the task is to find the maximum component size in the graph after each edge is added while constructing the graph.

Examples:

Input: N = 4, arr[][] = {{1, 2}, {3, 4}, {2, 3}}
Output: 2 2 4
Explanation:
Initially, the graph has 4 individual nodes 1, 2, 3 and 4.
After the first edge is added : 1 – 2, 3, 4 -> maximum component size = 2
After the second edge is added : 1 – 2, 3 – 4 -> maximum component size = 2
After the third edge is added : 1 – 2 – 3 – 4 -> maximum component size = 4

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: The naive approach for this problem is to add the edges sequentially and at each step apply depth-first search algorithm to find the size of the largest component.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the ` `// maximum comake_paironent size ` `// after addition of each ` `// edge to the graph ` `#include ` ` `  `using` `namespace` `std; ` ` `  `// Function to perform ` `// Depth First Search ` `// on the given graph ` `int` `dfs(``int` `u, ``int` `visited[], ` `        ``vector<``int``>* adj) ` `{ ` `    ``// Mark visited ` `    ``visited[u] = 1; ` `    ``int` `size = 1; ` ` `  `    ``// Add each child's ` `    ``// comake_paironent size ` `    ``for` `(``auto` `child : adj[u]) { ` `        ``if` `(!visited[child]) ` `            ``size += dfs(child, ` `                        ``visited, adj); ` `    ``} ` `    ``return` `size; ` `} ` ` `  `// Function to find the maximum ` `// comake_paironent size ` `// after addition of each ` `// edge to the graph ` `void` `maxSize(vector > e, ` `             ``int` `n) ` `{ ` `    ``// Graph in the adjacency ` `    ``// list format ` `    ``vector<``int``> adj[n]; ` ` `  `    ``// Visited array ` `    ``int` `visited[n]; ` ` `  `    ``vector<``int``> answer; ` ` `  `    ``// At each step, add a new ` `    ``// edge and apply dfs on all ` `    ``// the nodes to find the maximum ` `    ``// comake_paironent size ` `    ``for` `(``auto` `edge : e) { ` ` `  `        ``// Add this edge to undirected graph ` `        ``adj[edge.first - 1].push_back( ` `            ``edge.second - 1); ` `        ``adj[edge.second - 1].push_back( ` `            ``edge.first - 1); ` ` `  `        ``// Mark all the nodes ` `        ``// as unvisited ` `        ``memset``(visited, 0, ` `               ``sizeof``(visited)); ` ` `  `        ``int` `maxAns = 0; ` ` `  `        ``// Loop to perform DFS ` `        ``// and find the size ` `        ``// of the maximum comake_paironent ` `        ``for` `(``int` `i = 0; i < n; i++) { ` `            ``if` `(!visited[i]) { ` `                ``maxAns = max(maxAns, ` `                             ``dfs(i, visited, adj)); ` `            ``} ` `        ``} ` `        ``answer.push_back(maxAns); ` `    ``} ` ` `  `    ``// Print the answer ` `    ``for` `(``auto` `i : answer) { ` `        ``cout << i << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 4; ` `    ``vector > E; ` `    ``E.push_back(make_pair(1, 2)); ` `    ``E.push_back(make_pair(3, 4)); ` `    ``E.push_back(make_pair(2, 3)); ` ` `  `    ``maxSize(E, N); ` ` `  `    ``return` `0; ` `} `

Output:

```2 2 4
```

Time Complexity: O(|E| * N)

Efficient Approach: The idea is to use the concept of Disjoint Set (Union by rank and Path compression) to solve the problem more efficiently.

• Each node is initially a disjoint set within itself. As and when the edges are added, the disjoint sets are merged together forming larger components. In the disjoint set implementation, we will make the ranking system based on component sizes i.e when merging of two components is performed the larger component’s root will be considered the final root after the merge operation.
• One way to find the largest size component after each edge addition is to traverse the size array (size[i] represents the size of the component in which node ‘i’ belongs), but this is inefficient when the number of nodes in the graph is high.
• A more efficient way is to store the component sizes of all the root in some ordered data structure like sets.
• When two components are merged, all we need to do is remove the previous component sizes from the set and add the combined component size. So at each step, we would be able to find the largest component size in logarithmic complexity.

Below is the implementation of the above approach:

 `// C++ implementation to find the maximum ` `// component size after the addition of ` `// each edge to the graph ` ` `  `#include ` ` `  `using` `namespace` `std; ` ` `  `// Variables for implementing DSU ` `int` `par; ` `int` `size; ` ` `  `// Root of the component of node i ` `int` `root(``int` `i) ` `{ ` `    ``if` `(par[i] == i) ` `        ``return` `i; ` ` `  `    ``// Finding the root and applying ` `    ``// path compression ` `    ``else` `        ``return` `par[i] = root(par[i]); ` `} ` ` `  `// Function to merge two components ` `void` `merge(``int` `a, ``int` `b) ` `{ ` ` `  `    ``// Find the roots of both ` `    ``// the components ` `    ``int` `p = root(a); ` `    ``int` `q = root(b); ` ` `  `    ``// If both the nodes already belong ` `    ``// to the same compenent ` `    ``if` `(p == q) ` `        ``return``; ` ` `  `    ``// Union by rank, the rank in ` `    ``// this case is the size of ` `    ``// the component. ` `    ``// Smaller size will be ` `    ``// merged into larger, ` `    ``// so the larger's root ` `    ``// will be the final root ` `    ``if` `(size[p] > size[q]) ` `        ``swap(p, q); ` ` `  `    ``par[p] = q; ` `    ``size[q] += size[p]; ` `} ` ` `  `// Function to find the ` `// maximum component size ` `// after the addition of ` `// each edge to the graph ` `void` `maxSize(vector > e, ``int` `n) ` `{ ` ` `  `    ``// Initialising the disjoint set ` `    ``for` `(``int` `i = 1; i < n + 1; i++) { ` ` `  `        ``// Each node is the root and ` `        ``// each component size is 1 ` `        ``par[i] = i; ` `        ``size[i] = 1; ` `    ``} ` ` `  `    ``vector<``int``> answer; ` ` `  `    ``// A multiset is being used to store ` `    ``// the size of the components ` `    ``// because multiple components ` `    ``// can have same sizes ` `    ``multiset<``int``> compSizes; ` `    ``for` `(``int` `i = 1; i <= n; i++) ` `        ``compSizes.insert(size[i]); ` ` `  `    ``// At each step; add a new edge, ` `    ``// merge the components ` `    ``// and find the max ` `    ``// sized component ` `    ``for` `(``auto` `edge : e) { ` ` `  `        ``// Merge operation is required only when ` `        ``// both the nodes don't belong to the ` `        ``// same component ` `        ``if` `(root(edge.first) != root(edge.second)) { ` ` `  `            ``// Sizes of the compenents ` `            ``int` `size1 = size[root(edge.first)]; ` `            ``int` `size2 = size[root(edge.second)]; ` ` `  `            ``// Remove the previous component sizes ` `            ``compSizes.erase(compSizes.find(size1)); ` `            ``compSizes.erase(compSizes.find(size2)); ` ` `  `            ``// Perform the merge operation ` `            ``merge(edge.first, edge.second); ` ` `  `            ``// Insert the combined size ` `            ``compSizes.insert(size1 + size2); ` `        ``} ` ` `  `        ``// Maximum value in the multiset is ` `        ``// the max component size ` `        ``answer.push_back(*compSizes.rbegin()); ` `    ``} ` ` `  `    ``// Printing the answer ` `    ``for` `(``int` `i = 0; i < answer.size(); i++) { ` `        ``cout << answer[i] << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 4; ` `    ``vector > E; ` `    ``E.push_back(make_pair(1, 2)); ` `    ``E.push_back(make_pair(3, 4)); ` `    ``E.push_back(make_pair(2, 3)); ` ` `  `    ``maxSize(E, N); ` ` `  `    ``return` `0; ` `} `

Output:

```2 2 4
```

Time Complexity: O(|E| * log(N)) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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