Count the nodes in the given tree whose weight is even
Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is even.
Examples:
Input:
Output: 3
Only the weights of the nodes 2, 4 and 5 are even.
Approach: Perform dfs on the tree and for every node, check if it’s weight is divisible by 2 or not. If yes then increment count.
Algorithm:
Step 1: Start
Step 2: Create a global integer variable name it as “ans” and initialize it to 0.
Step 3: Create a vector of integer type name it as “weight” to store the weights of the nodes.
Step 4: Create a vector of integer type name it as “graph” to represent the tree.
Step 5: Create a static function of void return type name it as “dfs” which take node and its parent as input.
Step 6: The value of ans should be increased by one if the weight of the current node is even.
Step 7: Start a for loop for all graph node
a. if current node equal to parent then Recursively call the dfs function with the current node and the next node as n parameters.
Step 8: End
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int ans = 0;
vector< int > graph[100];
vector< int > weight(100);
void dfs( int node, int parent)
{
if (weight[node] % 2 == 0)
ans += 1;
for ( int to : graph[node]) {
if (to == parent)
continue ;
dfs(to, node);
}
}
int main()
{
int x = 15;
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
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
import java.util.*;
class GFG
{
static int ans = 0 ;
@SuppressWarnings ( "unchecked" )
static Vector<Integer>[] graph = new Vector[ 100 ];
static int [] weight = new int [ 100 ];
static void dfs( int node, int parent)
{
if (weight[node] % 2 == 0 )
ans += 1 ;
for ( int to : graph[node])
{
if (to == parent)
continue ;
dfs(to, node);
}
}
public static void main(String[] args)
{
int x = 15 ;
for ( int i = 0 ; i < 100 ; i++)
graph[i] = new Vector<>();
weight[ 1 ] = 5 ;
weight[ 2 ] = 10 ;
weight[ 3 ] = 11 ;
weight[ 4 ] = 8 ;
weight[ 5 ] = 6 ;
graph[ 1 ].add( 2 );
graph[ 2 ].add( 3 );
graph[ 2 ].add( 4 );
graph[ 1 ].add( 5 );
dfs( 1 , 1 );
System.out.println(ans);
}
}
|
Python3
ans = 0
graph = [[] for i in range ( 100 )]
weight = [ 0 ] * 100
def dfs(node, parent):
global ans
if (weight[node] % 2 = = 0 ):
ans + = 1
for to in graph[node]:
if (to = = parent):
continue
dfs(to, node)
x = 15
weight[ 1 ] = 5
weight[ 2 ] = 10
weight[ 3 ] = 11
weight[ 4 ] = 8
weight[ 5 ] = 6
graph[ 1 ].append( 2 )
graph[ 2 ].append( 3 )
graph[ 2 ].append( 4 )
graph[ 1 ].append( 5 )
dfs( 1 , 1 )
print (ans)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int ans = 0;
static List< int >[] graph = new List< int >[100];
static int [] weight = new int [100];
static void dfs( int node, int parent)
{
if (weight[node] % 2 == 0)
ans += 1;
foreach ( int to in graph[node])
{
if (to == parent)
continue ;
dfs(to, node);
}
}
public static void Main(String[] args)
{
for ( int i = 0; i < 100; i++)
graph[i] = new List< int >();
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
dfs(1, 1);
Console.WriteLine(ans);
}
}
|
Javascript
<script>
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 dfs(node, parent)
{
if (weight[node] % 2 == 0)
{
ans += 1;
}
for (let to=0;to<graph[node].length ;to++)
{
if (graph[node][to] == parent)
{
continue ;
}
dfs(graph[node][to], node);
}
}
let x = 15;
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
dfs(1, 1);
document.write(ans);
</script>
|
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) if 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.
Last Updated :
27 Apr, 2023
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