Dado un árbol, y los pesos de todos los Nodes, la tarea es contar el número de Nodes cuyo peso es un Cuadrado perfecto.
Ejemplos:
Aporte:
Salida: 3
Solo los pesos de los Nodes 1, 4 y 5 son cuadrados perfectos.
Enfoque: Realice dfs en el árbol y para cada Node, verifique si su peso es un cuadrado perfecto o no.
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 that returns true // if n is a perfect square bool isPerfectSquare(int n) { double x = sqrt(n); if (floor(x) != ceil(x)) return false; return true; } // Function to perform dfs void dfs(int node, int parent) { // If weight of the current node // is a perfect square if (isPerfectSquare(weight[node])) 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] = 4; weight[2] = 5; weight[3] = 3; weight[4] = 25; weight[5] = 16; weight[6] = 30; // Edges of the tree graph[1].push_back(2); graph[2].push_back(3); graph[2].push_back(4); graph[1].push_back(5); graph[5].push_back(6); dfs(1, 1); cout << ans; return 0; }
Java
// Java implementation of the approach import java.util.*; class GFG{ static int ans = 0; static Vector<Integer>[] graph = new Vector[100]; static int[] weight = new int[100]; // Function that returns true // if n is a perfect square static boolean isPerfectSquare(int n) { double x = Math.sqrt(n); if (Math.floor(x) != Math.ceil(x)) return false; return true; } // Function to perform dfs static void dfs(int node, int parent) { // If weight of the current node // is a perfect square if (isPerfectSquare(weight[node])) 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] = 4; weight[2] = 5; weight[3] = 3; weight[4] = 25; weight[5] = 16; weight[6] = 30; // Edges of the tree graph[1].add(2); graph[2].add(3); graph[2].add(4); graph[1].add(5); graph[5].add(6); dfs(1, 1); System.out.print(ans); } } // This code is contributed by Rajput-Ji
Python3
# Python3 implementation of the approach from math import * ans = 0 graph = [[] for i in range(100)] weight = [0] * 100 # Function that returns true # if n is a perfect square def isPerfectSquare(n): x = sqrt(n) if (floor(x) != ceil(x)): return False return True # Function to perform dfs def dfs(node, parent): global ans # If weight of the current node # is a perfect square if (isPerfectSquare(weight[node])): ans += 1; for to in graph[node]: if (to == parent): continue dfs(to, node) # Driver code x = 15 # Weights of the node weight[1] = 4 weight[2] = 5 weight[3] = 3 weight[4] = 25 weight[5] = 16 weight[6] = 30 # Edges of the tree graph[1].append(2) graph[2].append(3) graph[2].append(4) graph[1].append(5) graph[5].append(6) dfs(1, 1) print(ans) # This code is contributed by SHUBHAMSINGH10
C#
// C# program for the above approach using System; using System.Collections; using System.Collections.Generic; using System.Text; class GFG{ static int ans = 0; static ArrayList[] graph = new ArrayList[100]; static int[] weight = new int[100]; // Function that returns true // if n is a perfect square static bool isPerfectSquare(int n) { double x = Math.Sqrt(n); if (Math.Floor(x) != Math.Ceiling(x)) return false; return true; } // Function to perform dfs static void dfs(int node, int parent) { // If weight of the current node // is a perfect square if (isPerfectSquare(weight[node])) ans += 1; foreach(int to in 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 ArrayList(); // Weights of the node weight[1] = 4; weight[2] = 5; weight[3] = 3; weight[4] = 25; weight[5] = 16; weight[6] = 30; // Edges of the tree graph[1].Add(2); graph[2].Add(3); graph[2].Add(4); graph[1].Add(5); graph[5].Add(6); dfs(1, 1); Console.Write(ans); } } // This code is contributed by rutvik_56
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 n is a perfect square function isPerfectSquare(n) { let x = Math.sqrt(n); if (Math.floor(x) != Math.ceil(x)) return false; return true; } // Function to perform dfs function dfs(node,parent) { // If weight of the current node // is a perfect square if (isPerfectSquare(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 x = 15; // Weights of the node weight[1] = 4; weight[2] = 5; weight[3] = 3; weight[4] = 25; weight[5] = 16; weight[6] = 30; // Edges of the tree graph[1].push(2); graph[2].push(3); graph[2].push(4); graph[1].push(5); graph[5].push(6); dfs(1, 1); document.write( ans); // This code is contributed by unknown2108 </script>
Producción:
3
Análisis de Complejidad:
- Complejidad temporal: O(N*logV) donde V es el peso máximo de un Node en el árbol.
En DFS, cada Node del árbol se procesa una vez y, por lo tanto, la complejidad debida al DFS es O(N) para N Nodes en el árbol. Además, mientras se procesa cada Node, para verificar si el valor del Node es un cuadrado perfecto o no, se llama a la raíz cuadrada (V) incorporada, donde V es el peso del Node y esta función tiene una complejidad de O (log V). Por lo tanto, para cada Node, existe una complejidad adicional de O (log V). Por lo tanto, la complejidad temporal total es O(N*logV). - 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