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Queries to count connected components after removal of a vertex from a Tree
• Last Updated : 26 May, 2021

Given a Tree consisting of N nodes valued in the range [0, N) and an array Queries[] of Q integers consisting of values in the range [0, N). The task for each query is to remove the vertex valued Q[i] and count the connected components in the resulting graph.

Examples:

Input: N = 7, Edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}}, Queries[] = {0, 1, 6}

Output:
3 3 1
Explanation:
Query 1: Removal of the node valued 0 leads to removal of the edges (0, 1), (0, 2) and (0, 3). Therefore, the remaining graph has 3 connected components: [1, 4, 5], [2], [3, 6]
Query 2:Removal of the node valued 1 leads to removal of the edges (1, 4), (1, 5) and (1, 0). Therefore, remaining graph has 3 connected components: [4], [5], [2, 0, 3, 6]
Query 3:Removal of the node valued 6 leads to removal of the edges (3, 6). Therefore, the remaining graph has 1 connected component: [0, 1, 2, 3, 4, 5].

Input: N = 7, Edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}}, Queries[] = {5, 3, 2}

Output:1 2 1
Explanation:
Query 1: Removal of the node valued 5 leads to removal of the edge (1, 5). Therefore, the remaining graph has 1 connected component: [0, 1, 2, 3, 4, 6]
Query 2:Removal of the node valued 3 leads to removal of the edges (0, 3), (3, 6). Therefore, the remaining graph has 2 connected components: [0, 1, 2, 4, 5], [6]
Query 3: Removal of the node valued 2 leads to removal of the edge (0, 2). Therefore, the remaining graph has 1 connected component: [0, 1, 3, 4, 5, 6]

Approach: The idea is to observe that in a Tree, whenever a node is deleted, the nodes which were connected together to that node get separated. So, the count of connected components becomes equal to the degree of the deleted node.
Therefore, the approach is to precompute and store the degree of each node in an array. For every query, the count of the connected components is simply the degree of the corresponding node in the query.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;``#define MAX 100005` `// Stores degree of the nodes``int` `degree[MAX];` `// Function that finds the degree of``// each node in the given graph``void` `DegreeOfNodes(``int` `Edges[][2],``                   ``int` `N)``{``    ``// Precompute degrees of each node``    ``for` `(``int` `i = 0; i < N - 1; i++) {``        ``degree[Edges[i][0]]++;``        ``degree[Edges[i][1]]++;``    ``}``}` `// Function to print the number of``// connected components``void` `findConnectedComponents(``int` `x)``{``    ``// Print the degree of node x``    ``cout << degree[x] << ``' '``;``}` `// Function that counts the connected``// components after removing a vertex``// for each query``void` `countCC(``int` `N, ``int` `Q, ``int` `Queries[],``             ``int` `Edges[][2])``{` `    ``// Count degree of each node``    ``DegreeOfNodes(Edges, N);` `    ``// Interate over each query``    ``for` `(``int` `i = 0; i < Q; i++) {` `        ``// Find connected components``        ``// after removing given vertex``        ``findConnectedComponents(Queries[i]);``    ``}``}` `// Driver Code``int` `main()``{``    ``// Given N nodes and Q queries``    ``int` `N = 7, Q = 3;` `    ``// Given array of queries``    ``int` `Queries[] = { 0, 1, 6 };` `    ``// Given Edges``    ``int` `Edges[][2] = { { 0, 1 }, { 0, 2 },``                       ``{ 0, 3 }, { 1, 4 },``                       ``{ 1, 5 }, { 3, 6 } };` `    ``// Function Call``    ``countCC(N, Q, Queries, Edges);` `    ``return` `0;``}`

## Java

 `// Java program for``// the above approach``import` `java.util.*;``class` `GFG{``  ` `static` `final` `int` `MAX  = ``100005``;` `// Stores degree of the nodes``static` `int` `[]degree = ``new` `int``[MAX];` `// Function that finds the degree of``// each node in the given graph``static` `void` `DegreeOfNodes(``int` `[][]Edges,``                          ``int` `N)``{``  ``// Precompute degrees of each node``  ``for` `(``int` `i = ``0``; i < N - ``1``; i++)``  ``{``    ``degree[Edges[i][``0``]]++;``    ``degree[Edges[i][``1``]]++;``  ``}``}` `// Function to print the number of``// connected components``static` `void` `findConnectedComponents(``int` `x)``{``  ``// Print the degree of node x``  ``System.out.print(degree[x] + ``" "``);``}` `// Function that counts the connected``// components after removing a vertex``// for each query``static` `void` `countCC(``int` `N, ``int` `Q,``                    ``int` `Queries[],``                    ``int` `[][]Edges)``{``  ``// Count degree of each node``  ``DegreeOfNodes(Edges, N);` `  ``// Interate over each query``  ``for` `(``int` `i = ``0``; i < Q; i++)``  ``{``    ``// Find connected components``    ``// after removing given vertex``    ``findConnectedComponents(Queries[i]);``  ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ``// Given N nodes and Q queries``  ``int` `N = ``7``, Q = ``3``;` `  ``// Given array of queries``  ``int` `Queries[] = {``0``, ``1``, ``6``};` `  ``// Given Edges``  ``int` `[][]Edges = {{``0``, ``1``}, {``0``, ``2``},``                   ``{``0``, ``3``}, {``1``, ``4``},``                   ``{``1``, ``5``}, {``3``, ``6``}};` `  ``// Function Call``  ``countCC(N, Q, Queries, Edges);``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program for the above approach``MAX` `=` `100005` `# Stores degree of the nodes``degree ``=` `[``0``] ``*` `MAX` `# Function that finds the degree of``# each node in the given graph``def` `DegreeOfNodes(Edges, N):``    ` `    ``# Precompute degrees of each node``    ``for` `i ``in` `range``(N ``-` `1``):``        ``degree[Edges[i][``0``]] ``+``=` `1``        ``degree[Edges[i][``1``]] ``+``=` `1` `# Function to prthe number of``# connected components``def` `findConnectedComponents(x):``    ` `    ``# Print the degree of node x``    ``print``(degree[x], end ``=` `" "``)` `# Function that counts the connected``# components after removing a vertex``# for each query``def` `countCC(N, Q, Queries, Edges):` `    ``# Count degree of each node``    ``DegreeOfNodes(Edges, N)` `    ``# Interate over each query``    ``for` `i ``in` `range``(Q):` `        ``# Find connected components``        ``# after removing given vertex``        ``findConnectedComponents(Queries[i])` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given N nodes and Q queries``    ``N ``=` `7``    ``Q ``=` `3` `    ``# Given array of queries``    ``Queries ``=` `[ ``0``, ``1``, ``6` `]` `    ``# Given Edges``    ``Edges ``=` `[ [ ``0``, ``1` `], [ ``0``, ``2` `],``              ``[ ``0``, ``3` `], [ ``1``, ``4` `],``              ``[ ``1``, ``5` `], [ ``3``, ``6` `] ]` `    ``# Function call``    ``countCC(N, Q, Queries, Edges)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for``// the above approach``using` `System;``class` `GFG{``  ` `static` `readonly` `int` `MAX  = 100005;` `// Stores degree of the nodes``static` `int` `[]degree = ``new` `int``[MAX];` `// Function that finds the degree of``// each node in the given graph``static` `void` `DegreeOfNodes(``int` `[,]Edges,``                          ``int` `N)``{``  ``// Precompute degrees of each node``  ``for` `(``int` `i = 0; i < N - 1; i++)``  ``{``    ``degree[Edges[i, 0]]++;``    ``degree[Edges[i, 1]]++;``  ``}``}` `// Function to print the number of``// connected components``static` `void` `findConnectedComponents(``int` `x)``{``  ``// Print the degree of node x``  ``Console.Write(degree[x] + ``" "``);``}` `// Function that counts the connected``// components after removing a vertex``// for each query``static` `void` `countCC(``int` `N, ``int` `Q,``                    ``int` `[]Queries,``                    ``int` `[,]Edges)``{``  ``// Count degree of each node``  ``DegreeOfNodes(Edges, N);` `  ``// Interate over each query``  ``for` `(``int` `i = 0; i < Q; i++)``  ``{``    ``// Find connected components``    ``// after removing given vertex``    ``findConnectedComponents(Queries[i]);``  ``}``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ``// Given N nodes and Q queries``  ``int` `N = 7, Q = 3;` `  ``// Given array of queries``  ``int` `[]Queries = {0, 1, 6};` `  ``// Given Edges``  ``int` `[,]Edges = {{0, 1}, {0, 2},``                  ``{0, 3}, {1, 4},``                  ``{1, 5}, {3, 6}};` `  ``// Function Call``  ``countCC(N, Q, Queries, Edges);``}``}` `// This code is contributed by shikhasingrajput`

## Javascript

 ``
Output:
`3 3 1`

Time Complexity: O(E + Q), where E is the number of edges(E = N – 1) and Q is the number of queries.
Auxiliary Space: O(V), where V is number the of vertices.

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