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