Count the nodes in the given tree whose weight is prime

Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is prime.

**Examples:**

Input:

Output:2

Only the weights of the nodes 1 and 3 are prime.

**Approach:** Perform dfs on the tree and for every node, check if it’s weight is prime or not.

Below is the implementation of above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `int` `ans = 0; ` ` ` `vector<` `int` `> graph[100]; ` `vector<` `int` `> weight(100); ` ` ` `// Function that returns true ` `// if n is prime ` `bool` `isprime(` `int` `n) ` `{ ` ` ` `for` `(` `int` `i = 2; i * i <= n; i++) ` ` ` `if` `(n % i == 0) ` ` ` `return` `false` `; ` ` ` `return` `true` `; ` `} ` ` ` `// Function to perform dfs ` `void` `dfs(` `int` `node, ` `int` `parent) ` `{ ` ` ` `// If weight of node is prime or not ` ` ` `if` `(isprime(weight[node])) ` ` ` `ans += 1; ` ` ` ` ` `for` `(` `int` `to : graph[node]) { ` ` ` `if` `(to == parent) ` ` ` `continue` `; ` ` ` `dfs(to, node); ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `// 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` `int` `ans = ` `0` `; ` ` ` `static` `Vector<Integer>[] graph = ` `new` `Vector[` `100` `]; ` `static` `int` `[] weight = ` `new` `int` `[` `100` `]; ` ` ` `// Function that returns true ` `// if n is prime ` `static` `boolean` `isprime(` `int` `n) ` `{ ` ` ` `for` `(` `int` `i = ` `2` `; i * i <= n; i++) ` ` ` `if` `(n % i == ` `0` `) ` ` ` `return` `false` `; ` ` ` `return` `true` `; ` `} ` ` ` `// Function to perform dfs ` `static` `void` `dfs(` `int` `node, ` `int` `parent) ` `{ ` ` ` `// If weight of node is prime or not ` ` ` `if` `(isprime(weight[node])) ` ` ` `ans += ` `1` `; ` ` ` ` ` `for` `(` `int` `to : graph[node]) { ` ` ` `if` `(to == parent) ` ` ` `continue` `; ` ` ` `dfs(to, node); ` ` ` `} ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `for` `(` `int` `i = ` `0` `; i < ` `100` `; i++) ` ` ` `graph[i] = ` `new` `Vector<>(); ` ` ` ` ` `// 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` `].add(` `2` `); ` ` ` `graph[` `2` `].add(` `3` `); ` ` ` `graph[` `2` `].add(` `4` `); ` ` ` `graph[` `1` `].add(` `5` `); ` ` ` ` ` `dfs(` `1` `, ` `1` `); ` ` ` ` ` `System.out.print(ans); ` `} ` `} ` ` ` `// This code is contributed by Rajput-Ji ` |

## Python3

`# Python3 implementation of the approach ` `ans ` `=` `0` ` ` `graph ` `=` `[[] ` `for` `i ` `in` `range` `(` `100` `)] ` `weight ` `=` `[` `0` `] ` `*` `100` ` ` `# Function that returns true ` `# if n is prime ` `def` `isprime(n): ` ` ` `i ` `=` `2` ` ` `while` `(i ` `*` `i <` `=` `n): ` ` ` `if` `(n ` `%` `i ` `=` `=` `0` `): ` ` ` `return` `False` ` ` `i ` `+` `=` `1` ` ` `return` `True` ` ` `# Function to perform dfs ` `def` `dfs(node, parent): ` ` ` `global` `ans ` ` ` ` ` `# If weight of the current node is even ` ` ` `if` `(isprime(weight[node])): ` ` ` `ans ` `+` `=` `1` `; ` ` ` ` ` `for` `to ` `in` `graph[node]: ` ` ` `if` `(to ` `=` `=` `parent): ` ` ` `continue` ` ` `dfs(to, node) ` ` ` `# Driver code ` ` ` `# 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; ` `using` `System.Collections.Generic; ` `using` `System.Text; ` ` ` `class` `GFG{ ` ` ` `static` `int` `ans = 0; ` `static` `ArrayList[] graph = ` `new` `ArrayList[100]; ` `static` `int` `[] weight = ` `new` `int` `[100]; ` ` ` `// Function that returns true ` `// if n is prime ` `static` `bool` `isprime(` `int` `n) ` `{ ` ` ` `for` `(` `int` `i = 2; i * i <= n; i++) ` ` ` `if` `(n % i == 0) ` ` ` `return` `false` `; ` ` ` ` ` `return` `true` `; ` `} ` ` ` `// Function to perform dfs ` `static` `void` `dfs(` `int` `node, ` `int` `parent) ` `{ ` ` ` ` ` `// If weight of node is prime or not ` ` ` `if` `(isprime(weight[node])) ` ` ` `ans += 1; ` ` ` ` ` `foreach` `(` `int` `to ` `in` `graph[node]) ` ` ` `{ ` ` ` `if` `(to == parent) ` ` ` `continue` `; ` ` ` ` ` `dfs(to, node); ` ` ` `} ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main(` `string` `[] args) ` `{ ` ` ` `for` `(` `int` `i = 0; i < 100; i++) ` ` ` `graph[i] = ` `new` `ArrayList(); ` ` ` ` ` `// 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].Add(2); ` ` ` `graph[2].Add(3); ` ` ` `graph[2].Add(4); ` ` ` `graph[1].Add(5); ` ` ` ` ` `dfs(1, 1); ` ` ` ` ` `Console.Write(ans); ` `} ` `} ` ` ` `// This code is contributed by rutvik_56 ` |

**Output:**

2

**Complexity Analysis:**

**Time Complexity:**O(N*sqrt(V)), where V is the maximum weight of a node in the given tree.

In DFS, every node of the tree is processed once and hence the complexity due to the DFS is O(N) when there are N total nodes in the tree. Also, while processing every node, in order to check if the node value is prime or not, a loop up to sqrt(V) is being run, where V is the weight of the node. Hence for every node, there is an added complexity of O(sqrt(V)). Therefore, the time complexity is O(N*sqrt(V)).**Auxiliary Space:**O(1).

Any extra space is not required, so the space complexity is constant.

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