# 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.

**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 <bits/stdc++.h>` `using` `namespace` `std;` `long` `ans = 0;` `int` `x;` `vector<` `int` `> graph[100];` `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[1] = 5;` ` ` `weight[2] = 10;` ` ` `weight[3] = 11;` ` ` `weight[4] = 8;` ` ` `weight[5] = 6;` ` ` `// Edges of the tree` ` ` `graph[1].push_back(2);` ` ` `graph[2].push_back(3);` ` ` `graph[2].push_back(4);` ` ` `graph[1].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<Vector<Integer>> graph=` `new` `Vector<Vector<Integer>>(); ` `static` `Vector<Integer> weight=` `new` `Vector<Integer>(); ` `// 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<Integer>());` ` ` `// 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<List<` `int` `>> graph = ` `new` `List<List<` `int` `>>(); ` `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[1].Add(2); ` ` ` `graph[2].Add(3); ` ` ` `graph[2].Add(4); ` ` ` `graph[1].Add(5); ` ` ` `dfs(1, 1); ` ` ` `Console.WriteLine(ans); ` `}` `}` `// This code contributed by Rajput-Ji` |

## Javascript

`<script>` ` ` `// Javascript implementation of the approach` ` ` ` ` `let ans = 0;` `let x;` `let graph = ` `new` `Array(100);` `let weight = ` `new` `Array(100);` `for` `(let i = 0; i < 100; i++)` `{` ` ` `graph[i] = [];` ` ` `weight[i] = 0;` `}` `// Function to perform dfs` `function` `dfs(node, parent)` `{` ` ` `// If weight of the current node` ` ` `// is divisible by x` ` ` `if` `(weight[node] % x == 0)` ` ` `ans += 1;` ` ` `for` `(let to=0;to<graph[node].length;to++) {` ` ` `if` `(graph[node][to] == parent)` ` ` `continue` ` ` `dfs(graph[node][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].push(2);` ` ` `graph[2].push(3);` ` ` `graph[2].push(4);` ` ` `graph[1].push(5);` ` ` `dfs(1, 1);` ` ` `document.write(ans);` ` ` `// This code is contributed by Dharanendra L V.` ` ` `</script>` |

**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.