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.
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 even if (weight[node] % 2 == 0) ans += 1; for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); } } // Driver code int main() { int x = 15; // 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
// 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 even if (weight[node] % 2 == 0) ans += 1; for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); } } // Driver code public static void main(String[] args) { int x = 15; 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 shubhamsingh10 |
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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 # If weight of the current node is even if (weight[node] % 2 == 0): ans += 1 for to in graph[node]: if (to == parent): continue dfs(to, node) # Driver code x = 15 # Weights of the node weight[1] = 5weight[2] = 10weight[3] = 11weight[4] = 8weight[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 |
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C#
// C# implementation of the approach 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]; // Function to perform dfs static void dfs(int node, int parent) { // If weight of the current node is even if (weight[node] % 2 == 0) ans += 1; foreach (int to in 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 List<int>(); // 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); Console.WriteLine(ans); } } // This code is contributed by Rajput-Ji |
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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) 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.
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