Cuente los Nodes en el árbol dado cuyo peso es un número de fibonacci

Dado un árbol con los pesos de todos los Nodes, la tarea es contar el número de Nodes cuyo peso es un número de Fibonacci.
Ejemplos: 
 

Aporte: 
 

Salida: 2 
Explicación: 
Los Nodes que tienen pesos 5 y 8 son Nodes de Fibonacci. 
Aporte: 
 

Salida: 3 
Explicación: 
Los Nodes que tienen pesos 1, 3 y 8 son Nodes de Fibonacci. 
 

Enfoque: La idea es realizar un dfs en el árbol y para cada Node, verificar si el peso es un número de Fibonacci o no. 
 

1. Genere un hash que contenga todos los números de Fibonacci utilizando la programación dinámica .
2. Usando el recorrido de búsqueda primero en profundidad, recorra cada Node del árbol y verifique si el Node es un número de Fibonacci o no al verificar si ese elemento está presente en el hash precalculado o no.
3. Finalmente, imprima el número total de Nodes de Fibonacci.

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ program to count the number of nodes
// in the tree whose weight is a
// Fibonacci number
 
#include <bits/stdc++.h>
using namespace std;
 
const int sz = 1e5;
int ans = 0;
 
vector<int> graph[100];
vector<int> weight(100);
 
// To store all fibonacci numbers
set<int> fib;
 
// Function to generate fibonacci numbers using
// Dynamic Programming and create hash table
// to check Fibonacci numbers
void fibonacci()
{
    // Inserting the first two Fibonacci numbers
    // in the hash
    int prev = 0, curr = 1, len = 2;
    fib.insert(prev);
    fib.insert(curr);
 
    // Computing the Fibonacci numbers until
    // the maximum number and storing them
    // in the hash
    while (len <= sz) {
        int temp = curr + prev;
        fib.insert(temp);
        prev = curr;
        curr = temp;
        len++;
    }
}
 
// Function to perform dfs
void dfs(int node, int parent)
{
    // Check if the weight of the node
    // is a Fibonacci number or not
    if (fib.find(weight[node]) != fib.end())
        ans += 1;
 
    // Performing DFS to iterate the
    // remaining nodes
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
 
// Driver code
int main()
{
    // 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);
 
    // Generate fibonacci numbers
    fibonacci();
 
    // Call the dfs function to
    // traverse through the tree
    dfs(1, 1);
 
    cout << ans << endl;
 
    return 0;
}

// Java program to count the number of nodes
// in the tree whose weight is a
// Fibonacci number
import java.util.*;
 
class GFG{
  
static int sz = (int) 1e5;
static int ans = 0;
  
static Vector<Integer> []graph = new Vector[100];
static int []weight = new int[100];
  
// To store all fibonacci numbers
static HashSet<Integer> fib = new HashSet<Integer>();
  
// Function to generate fibonacci numbers using
// Dynamic Programming and create hash table
// to check Fibonacci numbers
static void fibonacci()
{
    // Inserting the first two Fibonacci numbers
    // in the hash
    int prev = 0, curr = 1, len = 2;
    fib.add(prev);
    fib.add(curr);
  
    // Computing the Fibonacci numbers until
    // the maximum number and storing them
    // in the hash
    while (len <= sz) {
        int temp = curr + prev;
        fib.add(temp);
        prev = curr;
        curr = temp;
        len++;
    }
}
  
// Function to perform dfs
static void dfs(int node, int parent)
{
    // Check if the weight of the node
    // is a Fibonacci number or not
    if (fib.contains(weight[node]))
        ans += 1;
  
    // Performing DFS to iterate the
    // remaining nodes
    for (int to : 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 Vector<Integer>();
    }
 
    // 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);
  
    // Generate fibonacci numbers
    fibonacci();
  
    // Call the dfs function to
    // traverse through the tree
    dfs(1, 1);
  
    System.out.print(ans +"\n");
  
}
}
 
// This code is contributed by Rajput-Ji

# Python 3 program to count the number of nodes
# in the tree whose weight is a
# Fibonacci number
sz = 1e5
ans = 0
 
graph = [[] for i in range(100)]
weight = [0 for i in range(100)]
 
# To store all fibonacci numbers
fib = set()
 
# Function to generate fibonacci numbers using
# Dynamic Programming and create hash table
# to check Fibonacci numbers
def fibonacci():
 
    # Inserting the first two Fibonacci numbers
    # in the hash
    prev = 0
    curr = 1
    len1 = 2
    fib.add(prev)
    fib.add(curr)
 
    # Computing the Fibonacci numbers until
    # the maximum number and storing them
    # in the hash
    while (len1 <= sz):
        temp = curr + prev
        fib.add(temp)
        prev = curr;
        curr = temp;
        len1 += 1
 
# Function to perform dfs
def dfs(node, parent):
    global ans
 
    # Check if the weight of the node
    # is a Fibonacci number or not
    if (weight[node] in fib):
        ans += 1
 
    # Performing DFS to iterate the
    # remaining nodes
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
 
# Driver code
if __name__ == '__main__':
    # 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)
 
    # Generate fibonacci numbers
    fibonacci()
 
    # Call the dfs function to
    # traverse through the tree
    dfs(1, 1)
 
    print(ans)
 
# This code is contributed by Surendra_Gangwar

// C# program to count the number of nodes
// in the tree whose weight is a
// Fibonacci number
using System;
using System.Collections.Generic;
 
public class GFG{
   
static int sz = (int) 1e5;
static int ans = 0;
   
static List<int> []graph = new List<int>[100];
static int []weight = new int[100];
   
// To store all fibonacci numbers
static HashSet<int> fib = new HashSet<int>();
   
// Function to generate fibonacci numbers using
// Dynamic Programming and create hash table
// to check Fibonacci numbers
static void fibonacci()
{
    // Inserting the first two Fibonacci numbers
    // in the hash
    int prev = 0, curr = 1, len = 2;
    fib.Add(prev);
    fib.Add(curr);
   
    // Computing the Fibonacci numbers until
    // the maximum number and storing them
    // in the hash
    while (len <= sz) {
        int temp = curr + prev;
        fib.Add(temp);
        prev = curr;
        curr = temp;
        len++;
    }
}
   
// Function to perform dfs
static void dfs(int node, int parent)
{
    // Check if the weight of the node
    // is a Fibonacci number or not
    if (fib.Contains(weight[node]))
        ans += 1;
   
    // Performing DFS to iterate the
    // remaining nodes
    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);
   
    // Generate fibonacci numbers
    fibonacci();
   
    // Call the dfs function to
    // traverse through the tree
    dfs(1, 1);
   
    Console.Write(ans +"\n");
   
}
}
// This code contributed by Rajput-Ji

<script>
 
// JavaScript program to count the number of nodes
// in the tree whose weight is a
// Fibonacci number
 
var sz = 1000000;
var ans = 0;
   
var graph = Array.from(Array(100), ()=>Array());
var weight = Array(100);
   
// To store all fibonacci numbers
var fib = new Set();
   
// Function to generate fibonacci numbers using
// Dynamic Programming and create hash table
// to check Fibonacci numbers
function fibonacci()
{
    // Inserting the first two Fibonacci numbers
    // in the hash
    var prev = 0, curr = 1, len = 2;
    fib.add(prev);
    fib.add(curr);
   
    // Computing the Fibonacci numbers until
    // the maximum number and storing them
    // in the hash
    while (len <= sz) {
        var temp = curr + prev;
        fib.add(temp);
        prev = curr;
        curr = temp;
        len++;
    }
}
   
// Function to perform dfs
function dfs(node, parent)
{
    // Check if the weight of the node
    // is a Fibonacci number or not
    if (fib.has(weight[node]))
        ans += 1;
   
    // Performing DFS to iterate the
    // remaining nodes
    for(var to of graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
   
// Driver code
 
// 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);
 
// Generate fibonacci numbers
fibonacci();
 
// Call the dfs function to
// traverse through the tree
dfs(1, 1);
 
document.write(ans +"<br>");
 
</script>

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 a dfs es O(N) si hay un total de N Nodes en el árbol. Además, para procesar cada Node se utiliza la función fibonacci(), que también tiene una complejidad de O(N), pero dado que esta función se ejecuta solo una vez, no afecta la complejidad temporal general. Por lo tanto, la complejidad del tiempo es O(N).
• Espacio Auxiliar : O(N). 
  Se utiliza espacio extra para el hashset de Fibonacci, por lo que la complejidad del espacio es O(N).