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.