Count the nodes in the given tree whose weight is even parity

Given a tree and the weights of all the nodes, the task is to count the number of nodes whose weights are even parity i.e. whether the count of set bits in them is even.

Examples:

Input:

Output: 3



Weight Binary Representation Parity
5 0101 Even
10 1010 Even
11 1011 Odd
8 1000 Odd
6 0110 Even

Approach: Perform dfs on the tree and for every node, check if it’s weight is even parity or not. If yes then increment count.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
int ans = 0;
  
vector<int> graph[100];
vector<int> weight(100);
  
// Function that returns true if count
// of set bits in x is even
bool isEvenParity(int x)
{
    // parity will store the
    // count of set bits
    int parity = 0;
    while (x != 0) {
        x = x & (x - 1);
        parity++;
    }
  
    if (parity % 2 == 0)
        return true;
    else
        return false;
}
  
// Function to perform dfs
void dfs(int node, int parent)
{
    // If weight of the current
    // node has even parity
    if (isEvenParity(weight[node]))
        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;
}

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Java

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// Java implementation of the approach 
import java.util.*;
  
class GFG
{
  
static int ans = 0
  
static Vector<Vector<Integer>> graph = new Vector<Vector<Integer>>(); 
static Vector<Integer> weight = new Vector<Integer>(); 
  
// Function that returns true if count 
// of set bits in x is even 
static boolean isEvenParity(int x) 
    // parity will store the 
    // count of set bits 
    int parity = 0
    while (x != 0
    
        x = x & (x - 1); 
        parity++; 
    
  
    if (parity % 2 == 0
        return true
    else
        return false
  
// Function to perform dfs 
static void dfs(int node, int parent) 
    // If weight of the current 
    // node has even parity 
    if (isEvenParity(weight.get(node) ))
        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[])
    // 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

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C#

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// 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 that returns true if count 
// of set bits in x is even 
static bool isEvenParity(int x) 
    // parity will store the 
    // count of set bits 
    int parity = 0; 
    while (x != 0) 
    
        x = x & (x - 1); 
        parity++; 
    
  
    if (parity % 2 == 0) 
        return true
    else
        return false
  
// Function to perform dfs 
static void dfs(int node, int parent) 
    // If weight of the current 
    // node has even parity 
    if (isEvenParity(weight[node]))
        ans += 1; 
  
    for (int i = 0; i < graph[node].Count; i++) 
    
        if (graph[node][i] == parent) 
            continue
        dfs(graph[node][i] , node); 
    
  
// Driver code 
static void Main()
    // 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 mits

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Output:

3


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