# 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.

**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 <bits/stdc++.h>` `using` `namespace` `std;` `int` `ans = 0;` `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 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[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` `;` ` ` `@SuppressWarnings` `(` `"unchecked"` `)` ` ` `static` `Vector<Integer>[] 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<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 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[1].Add(2);` ` ` `graph[2].Add(3);` ` ` `graph[2].Add(4);` ` ` `graph[1].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` |

## Javascript

`<script>` ` ` `// Javascript implementation of the approach` ` ` `let ans = 0;` `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 a power of 2` ` ` `let x = weight[node];` ` ` `if` `(x && (!(x & (x - 1))))` ` ` `ans += 1;` ` ` `for` `(let to=0;to<graph[node].length;to++) {` ` ` `if` `(graph[node][to] == parent)` ` ` `continue` ` ` `dfs(graph[node][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].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:**

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.