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 its weight is even parity or not. If yes then increment count.
Steps to solve the problem:
- Initialize ans to 0.
- Define a function isEvenParity(x) that takes an integer x and returns true if the count of set bits in x is even and false otherwise.
- Define a function dfs(node, parent) that performs depth-first search on the graph.
- Within the dfs function:
a. Check if the weight of the current node has even parity using the isEvenParity function. If it does, increment ans by 1.
b. For each neighbor to of node in the graph, if it is not equal to the parent, recursively call dfs with to as the node and node as the parent. - Return ans as the final result.
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 that returns true if count// of set bits in x is evenbool 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 dfsvoid 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 codeint 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; 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 |
Python3
# Python3 implementation of the approachans = 0graph = [[] for i in range(100)]weight = [0]*100# Function that returns True if count# of set bits in x is evendef isEvenParity(x): # parity will store the # count of set bits parity = 0 while (x != 0): x = x & (x - 1) parity += 1 if (parity % 2 == 0): return True else: return False# Function to perform dfsdef dfs(node, parent): global ans # If weight of the current # node has even parity if (isEvenParity(weight[node])): ans += 1 for to in graph[node]: if (to == parent): continue dfs(to, node)# Driver code# Weights of the nodeweight[1] = 5weight[2] = 10weight[3] = 11weight[4] = 8weight[5] = 6# Edges of the treegraph[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 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 |
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 that returns true if count// of set bits in x is evenfunction isEvenParity(x){ // parity will store the // count of set bits let parity = 0; while (x != 0) { x = x & (x - 1); parity++; } if (parity % 2 == 0) return true; else return false;}// Function to perform dfsfunction dfs(node, parent){ // If weight of the current // node has even parity if (isEvenParity(weight[node])) 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:
3
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.
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