Dado un árbol y los pesos de todos los Nodes, la tarea es contar los Nodes cuyos pesos son divisibles por x .
Ejemplos:
Aporte:
x = 5
Salida: 2
Solo los Nodes 1 y 2 tienen pesos divisibles por 5.
Enfoque: Realice dfs en el árbol y para cada Node, verifique si su peso es divisible por x o no. Si es así, entonces incremente el conteo.
Implementación:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; long ans = 0; int x; 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 divisible by x if (weight[node] % x == 0) ans += 1; for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); } } // Driver code int main() { x = 5; // 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 long ans = 0; static int x; static Vector<Vector<Integer>> graph=new Vector<Vector<Integer>>(); static Vector<Integer> weight=new Vector<Integer>(); // Function to perform dfs static void dfs(int node, int parent) { // If weight of the current node // is divisible by x if (weight.get(node) % x == 0) 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[]) { x = 5; // 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 approach ans = 0 graph = [[] for i in range(100)] weight = [0] * 100 # Function to perform dfs def dfs(node, parent): global ans,x # If weight of the current node # is divisible by x if (weight[node] % x == 0): ans += 1 for to in graph[node]: if (to == parent): continue dfs(to, node) # Driver code x = 5 # 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 long ans = 0; static int x; static List<List<int>> graph = new List<List<int>>(); static List<int> weight = new List<int>(); // Function to perform dfs static void dfs(int node, int parent) { // If weight of the current node // is divisible by x if (weight[node] % x == 0) ans += 1; for (int i = 0; i < graph[node].Count; i++) { if (graph[node][i] == parent) continue; dfs(graph[node][i], node); } } // Driver code public static void Main(String []args) { x = 5; // 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 contributed by Rajput-Ji
Javascript
<script> // Javascript implementation of the approach let ans = 0; let x; 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 divisible by x if (weight[node] % x == 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 x = 5; // 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>
Producción:
2
Análisis de Complejidad:
- Complejidad temporal: O(N).
En DFS, cada Node del árbol se procesa una vez y, por lo tanto, la complejidad debida al DFS es O(N) cuando 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