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 dfsvoid 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 codeint 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 approachans = 0graph = [[] 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 nodeweight[1] = 5weight[2] = 10weight[3] = 11weight[4] = 8weight[5] = 6# Edges of the treegraph[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 dfsfunction 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.
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