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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 using namespace std;   const int sz = 1e5; int ans = 0;   vector graph[100]; vector weight(100);   // To store all fibonacci numbers set 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 []graph = new Vector[100]; static int []weight = new int[100];    // To store all fibonacci numbers static HashSet fib = new HashSet();    // 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();     }       // 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 []graph = new List[100]; static int []weight = new int[100];     // To store all fibonacci numbers static HashSet fib = new HashSet();     // 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();     }        // 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



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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