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# Count the nodes of the given tree whose weight has X as a factor

• Last Updated : 11 Jun, 2021

Given a tree, and the weights of all the nodes, the task is to count the nodes whose weights are divisible by x.
Examples:

Input:

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Output:
Only the nodes 1 and 2 have weights divisible by 5.

Approach: Perform dfs on the tree and for every node, check if it’s weight is divisible by x or not. If yes then increment the count.
Implementation:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `long` `ans = 0;``int` `x;``vector<``int``> graph;``vector<``int``> weight(100);` `// Function to perform dfs``void` `dfs(``int` `node, ``int` `parent)``{` `    ``// If weight of the current node``    ``// is divisible by x``    ``if` `(weight[node] % x == 0)``        ``ans += 1;` `    ``for` `(``int` `to : graph[node]) {``        ``if` `(to == parent)``            ``continue``;``        ``dfs(to, node);``    ``}``}` `// Driver code``int` `main()``{``    ``x = 5;` `    ``// Weights of the node``    ``weight = 5;``    ``weight = 10;``    ``weight = 11;``    ``weight = 8;``    ``weight = 6;` `    ``// Edges of the tree``    ``graph.push_back(2);``    ``graph.push_back(3);``    ``graph.push_back(4);``    ``graph.push_back(5);` `    ``dfs(1, 1);` `    ``cout << ans;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{``    ` `static` `long` `ans = ``0``;``static` `int` `x;``static` `Vector> graph=``new` `Vector>();``static` `Vector weight=``new` `Vector();` `// Function to perform dfs``static` `void` `dfs(``int` `node, ``int` `parent)``{` `    ``// If weight of the current node``    ``// is divisible by x``    ``if` `(weight.get(node) % x == ``0``)``        ``ans += ``1``;` `    ``for` `(``int` `i = ``0``; i < graph.get(node).size(); i++)``    ``{``        ``if` `(graph.get(node).get(i) == parent)``            ``continue``;``        ``dfs(graph.get(node).get(i), node);``    ``}``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``x = ``5``;` `    ``// Weights of the node``    ``weight.add(``0``);``    ``weight.add(``5``);``    ``weight.add(``10``);;``    ``weight.add(``11``);;``    ``weight.add(``8``);``    ``weight.add(``6``);``    ` `    ``for``(``int` `i = ``0``; i < ``100``; i++)``    ``graph.add(``new` `Vector());` `    ``// Edges of the tree``    ``graph.get(``1``).add(``2``);``    ``graph.get(``2``).add(``3``);``    ``graph.get(``2``).add(``4``);``    ``graph.get(``1``).add(``5``);` `    ``dfs(``1``, ``1``);` `    ``System.out.println(ans);``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 implementation of the approach``ans ``=` `0` `graph ``=` `[[] ``for` `i ``in` `range``(``100``)]``weight ``=` `[``0``] ``*` `100` `# Function to perform dfs``def` `dfs(node, parent):``    ``global` `ans,x``    ` `    ``# If weight of the current node``    ``# is divisible by x``    ``if` `(weight[node] ``%` `x ``=``=` `0``):``        ``ans ``+``=` `1``    ``for` `to ``in` `graph[node]:``        ``if` `(to ``=``=` `parent):``            ``continue``        ``dfs(to, node)` `# Driver code``x ``=` `5` `# Weights of the node``weight[``1``] ``=` `5``weight[``2``] ``=` `10``weight[``3``] ``=` `11``weight[``4``] ``=` `8``weight[``5``] ``=` `6` `# Edges of the tree``graph[``1``].append(``2``)``graph[``2``].append(``3``)``graph[``2``].append(``4``)``graph[``1``].append(``5``)` `dfs(``1``, ``1``)``print``(ans)` `# This code is contributed by SHUBHAMSINGH10`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ` `static` `long` `ans = 0;``static` `int` `x;``static` `List> graph = ``new` `List>();``static` `List<``int``> weight = ``new` `List<``int``>();` `// Function to perform dfs``static` `void` `dfs(``int` `node, ``int` `parent)``{` `    ``// If weight of the current node``    ``// is divisible by x``    ``if` `(weight[node] % x == 0)``        ``ans += 1;` `    ``for` `(``int` `i = 0; i < graph[node].Count; i++)``    ``{``        ``if` `(graph[node][i] == parent)``            ``continue``;``        ``dfs(graph[node][i], node);``    ``}``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``x = 5;` `    ``// Weights of the node``    ``weight.Add(0);``    ``weight.Add(5);``    ``weight.Add(10);;``    ``weight.Add(11);;``    ``weight.Add(8);``    ``weight.Add(6);``    ` `    ``for``(``int` `i = 0; i < 100; i++)``    ``graph.Add(``new` `List<``int``>());` `    ``// Edges of the tree``    ``graph.Add(2);``    ``graph.Add(3);``    ``graph.Add(4);``    ``graph.Add(5);` `    ``dfs(1, 1);` `    ``Console.WriteLine(ans);``}``}` `// This code contributed by Rajput-Ji`

## Javascript

 ``
Output:
`2`

Complexity Analysis:

• Time Complexity: O(N).
In DFS, every node of the tree is processed once and hence the complexity due to the DFS is O(N) when there are total N nodes in the tree. Therefore, the time complexity is O(N).
• Auxiliary Space: O(1).
Any extra space is not required, so the space complexity is constant.

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