# 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.

## 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 prime or not.

Below is the implementation of above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `int` `ans = 0; ` ` `  `vector<``int``> graph; ` `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 = 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` `int` `ans = ``0``; ` ` `  `static` `Vector[] 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 `

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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