# Count the nodes in the given tree whose weight is a power of two

Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is a power of 2.

Examples:

Input: Output: 1
Only the weight of the node 4 is a power of 2.

## 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 its weight is a power of 2 or not, if yes then increment the count.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `int` `ans = 0; ` ` `  `vector<``int``> graph; ` `vector<``int``> weight(100); ` ` `  `// Function to perform dfs ` `void` `dfs(``int` `node, ``int` `parent) ` `{ ` `    ``// If weight of the current node ` `    ``// is a power of 2 ` `    ``int` `x = weight[node]; ` `    ``if` `(x && (!(x & (x - 1)))) ` `        ``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``; ` ` `  `    ``@SuppressWarnings``(``"unchecked"``) ` `    ``static` `Vector[] graph = ``new` `Vector[``100``]; ` `    ``static` `int``[] weight = ``new` `int``[``100``]; ` ` `  `    ``// Function to perform dfs ` `    ``static` `void` `dfs(``int` `node, ``int` `parent) ` `    ``{ ` `        ``// If weight of the current node ` `        ``// is a power of 2 ` `        ``int` `x = weight[node]; ` `        ``if` `(x != ``0` `&& (x & (x - ``1``)) == ``0``) ` `            ``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.println(ans); ` `    ``} ` `} ` ` `  `// This code is contributed by ` `// sanjeev2552 `

## C#

 `// C# implementation of the approach  ` `using` `System;  ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG  ` `{  ` `     `  `static` `int` `ans = 0;  ` `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 a power of 2  ` `    ``int` `x = weight[node];  ` `    ``bool` `result = Convert.ToBoolean((x & (x - 1))); ` `    ``bool` `result1 = Convert.ToBoolean(x); ` `    ``if` `(result1 && (!result))  ` `        ``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)  ` `{  ` `    ``// 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 is contributed by shubhamsingh10 `

## 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` `mini, graph, weight, ans  ` `     `  `    ``# If weight of the current node  ` `    ``# is a power of 2  ` `    ``x ``=` `weight[node] ` `    ``if` `(x ``and` `(``not` `(x & (x ``-` `1``)))): ` `        ``ans ``+``=` `1` `    ``for` `to ``in` `graph[node]:  ` `        ``if` `(to ``=``=` `parent):  ` `            ``continue` `        ``dfs(to, node)  ` `         `  `        ``# Calculating the weighted  ` `        ``# sum of the subtree  ` `        ``weight[node] ``+``=` `weight[to]  ` `     `  `# 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:

```1
```

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) for 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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