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: 2
Explanation:
Nodes having weights 5 and 8 are fibonacci nodes.
Input:
Output: 3
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.
- Generate a hash containing all the Fibonacci numbers using Dynamic programming.
- 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.
- 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 numbersset<int> fib;// Function to generate fibonacci numbers using// Dynamic Programming and create hash table// to check Fibonacci numbersvoid 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 dfsvoid 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 codeint 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 numberimport 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 numbersstatic HashSet<Integer> fib = new HashSet<Integer>(); // Function to generate fibonacci numbers using// Dynamic Programming and create hash table// to check Fibonacci numbersstatic 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 dfsstatic 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 codepublic 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 numbersz = 1e5ans = 0graph = [[] for i in range(100)]weight = [0 for i in range(100)]# To store all fibonacci numbersfib = set()# Function to generate fibonacci numbers using# Dynamic Programming and create hash table# to check Fibonacci numbersdef 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 dfsdef 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 codeif __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 numberusing 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 numbersstatic HashSet<int> fib = new HashSet<int>(); // Function to generate fibonacci numbers using// Dynamic Programming and create hash table// to check Fibonacci numbersstatic 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 dfsstatic 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 codepublic 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 numbervar sz = 1000000;var ans = 0; var graph = Array.from(Array(100), ()=>Array());var weight = Array(100); // To store all fibonacci numbersvar fib = new Set(); // Function to generate fibonacci numbers using// Dynamic Programming and create hash table// to check Fibonacci numbersfunction 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 dfsfunction 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 nodeweight[1] = 5;weight[2] = 10;weight[3] = 11;weight[4] = 8;weight[5] = 6;// Edges of the treegraph[1].push(2);graph[2].push(3);graph[2].push(4);graph[1].push(5);// Generate fibonacci numbersfibonacci();// Call the dfs function to// traverse through the treedfs(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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