Dado un árbol y los pesos de todos los Nodes, la tarea es contar el número de Nodes cuyo peso es par.
Ejemplos:
Aporte:
Salida: 3
Solo los pesos de los Nodes 2, 4 y 5 son pares.
Enfoque: Realice dfs en el árbol y para cada Node, verifique si su peso es divisible por 2 o no. Si es así, entonces incremente el conteo.
A continuación se muestra la implementación del enfoque anterior:
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; }
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
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] = 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 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
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 to perform dfs function dfs(node, parent) { // If weight of the current node is even 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); } } // Driver code let 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(2); graph[2].push(3); graph[2].push(4); graph[1].push(5); dfs(1, 1); document.write(ans); // This code is contributed by unknown2108 </script>
Producción:
3
Análisis de Complejidad:
- Complejidad temporal: O(N).
En dfs, cada Node del árbol se procesa una vez y, por lo tanto, la complejidad debida a dfs es O(N) si hay un total de N Nodes en el árbol. Por lo tanto, la complejidad del tiempo es O(N). - Espacio Auxiliar : O(1).
No se requiere ningún espacio adicional, por lo que la complejidad del espacio es constante.
Publicación traducida automáticamente
Artículo escrito por mohit kumar 29 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA