Given a tree, and the weights of all the nodes, the task is to count the nodes whose weights are divisible by x.
Examples:
Input:
x = 5
Output: 2
Only the nodes 1 and 2 have weights divisible by 5.
Approach: Perform dfs on the tree and for every node, check if it’s weight is divisible by x or not. If yes then increment the count.
Implementation:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
long ans = 0;
int x;
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 divisible by x
if (weight[node] % x == 0)
ans += 1;
for ( int to : graph[node]) {
if (to == parent)
continue ;
dfs(to, node);
}
} // Driver code int main()
{ x = 5;
// 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 long ans = 0 ;
static int x;
static Vector<Vector<Integer>> graph= new Vector<Vector<Integer>>();
static Vector<Integer> weight= new Vector<Integer>();
// Function to perform dfs static void dfs( int node, int parent)
{ // If weight of the current node
// is divisible by x
if (weight.get(node) % x == 0 )
ans += 1 ;
for ( int i = 0 ; i < graph.get(node).size(); i++)
{
if (graph.get(node).get(i) == parent)
continue ;
dfs(graph.get(node).get(i), node);
}
} // Driver code public static void main(String args[])
{ x = 5 ;
// 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 Vector<Integer>());
// Edges of the tree
graph.get( 1 ).add( 2 );
graph.get( 2 ).add( 3 );
graph.get( 2 ).add( 4 );
graph.get( 1 ).add( 5 );
dfs( 1 , 1 );
System.out.println(ans);
} } // This code is contributed by Arnab Kundu |
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 ans,x
# If weight of the current node
# is divisible by x
if (weight[node] % x = = 0 ):
ans + = 1
for to in graph[node]:
if (to = = parent):
continue
dfs(to, node)
# Driver code x = 5
# 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 |
C#
// C# implementation of the approach using System;
using System.Collections.Generic;
class GFG
{ static long ans = 0;
static int x;
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 divisible by x
if (weight[node] % x == 0)
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)
{ x = 5;
// 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 contributed by Rajput-Ji |
Javascript
<script> // Javascript implementation of the approach let ans = 0; let x; 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 divisible by x
if (weight[node] % x == 0)
ans += 1;
for (let to=0;to<graph[node].length;to++) {
if (graph[node][to] == parent)
continue
dfs(graph[node][to], node);
}
} // Driver code x = 5;
// 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:
2
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) when there are total 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.