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Count nodes in the given tree whose weight is a fibonacci number

  • Last Updated : 23 Jun, 2021

Given a tree with the weights of all the nodes, the task is to count the number of nodes whose weight is a Fibonacci number.
Examples: 
 

Input: 
 

Output:
Explanation: 
Nodes having weights 5 and 8 are fibonacci nodes. 
Input: 
 



Output:
Explanation: 
Nodes having weights 1, 3 and 8 are fibonacci nodes. 
 

 

Approach: The idea is to perform a dfs on the tree and for every node, check whether the weight is a Fibonacci number or not. 
 

  1. Generate a hash containing all the Fibonacci numbers using Dynamic programming.
  2. Using depth-first search traversal, traverse through every node of the tree and check whether the node is a Fibonacci number or not by checking if that element is present in the precomputed hash or not.
  3. Finally, print the total number of Fibonacci nodes.

Below is the implementation of above approach: 
 

C++




// 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




// 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

Python3




# 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#




// 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

Javascript




<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>
Output: 
2

 

Complexity Analysis: 
 

  • Time Complexity : O(N). 
    In dfs, every node of the tree is processed once and hence the complexity due to the dfs is O(N) if there are total N nodes in the tree. Also, for processing each node the fibonacci() function is used which has a complexity of O(N) too but since this function is executed only once, it does not affect the overall time complexity. Therefore, the time complexity is O(N).
  • Auxiliary Space : O(N). 
    Extra space is used for the fibonacci hashset, so the space complexity is O(N).

 

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