# Count the nodes in the given tree whose weight is prime

• Last Updated : 20 Apr, 2021

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:
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 ` `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[] 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`

## Javascript

 ``

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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