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

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

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

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 `

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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