Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 11 Jun, 2021

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: 
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.



Output:
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.

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :