Given a tree, and the weights of all the nodes and an integer X, the task is to count all the nodes i such that (weight[i] + X) is a Fibonacci Number.
First, few Fibonacci numbers are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, …
Examples:
Input:
X = 5
Output: 2
Only the nodes 3 and 5 give a fibonacci number when 5 is added to them.
i.e. (3 + 5) = 8 and (16 + 5) = 21 are both Fibonacci numbers.
Approach: Perform dfs on the tree and count all the nodes sum of whose weight with x is a Fibonacci number.
Algorithm:
- Create a static variable “ans” and “x” of integer type.
- Create an ArrayList of type ‘ArrayList’ called “graph” and an ArrayList of type ‘integer’ called “weight”.
- Create a method called “isPerfectSquare” that takes a double type argument x and checks if the square root of x is an integer or not.
- Create a method called “isFibonacci” that takes an integer type argument n and checks if 5nn+4 or 5nn-4 is a perfect square or not.
- Create a method called “dfs” that takes two integer type arguments node and parent.
- In the “dfs” method, if the sum of the weight of the current node and x is a Fibonacci number, increment the ans variable by 1.
- Traverse through the adjacent nodes of the current node and call the “dfs” method recursively for each adjacent node if it is not equal to the parent node.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;int ans = 0, x;vector<int> graph[100];vector<int> weight(100);// Function that returns true if// x is a perfect squarebool isPerfectSquare(long double x){ // Find floating point value of // square root of x long double sr = sqrt(x); // If square root is an integer return ((sr - floor(sr)) == 0);}// Function that returns true// if n is a fibonacci numberbool isFibonacci(int n){ return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);}// Function to perform dfsvoid dfs(int node, int parent){ // If weight of the current node // gives a fibonacci number // when x is added to it if (isFibonacci(weight[node] + x)) ans += 1; for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); }}// Driver codeint main(){ x = 5; // Weights of the node weight[1] = 4; weight[2] = 5; weight[3] = 3; weight[4] = 25; weight[5] = 16; weight[6] = 34; // 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 // above approachimport java.util.*;@SuppressWarnings("unchecked")class GFG{ static int ans = 0, x; static ArrayList []graph = new ArrayList[100];static ArrayList weight = new ArrayList(); // Function that returns true if// x is a perfect squarestatic boolean isPerfectSquare(double x){ // Find floating point value of // square root of x double sr = Math.sqrt(x); // If square root is an integer return ((sr - Math.floor(sr)) == 0);} // Function that returns true// if n is a fibonacci numberstatic boolean isFibonacci(int n){ return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);} // Function to perform dfsstatic void dfs(int node, int parent){ // If weight of the current node // gives a fibonacci number // when x is added to it if (isFibonacci((int)weight.get(node) + x)) ans += 1; for(int to : (ArrayList<Integer>)graph[node]) { if (to == parent) continue; dfs(to, node); }} // Driver Codepublic static void main(String[] args){ x = 5; for(int i = 0; i < 100; i++) { weight.add(0); graph[i] = new ArrayList(); } // Weights of the node weight.add(1, 4); weight.add(2, 5); weight.add(3, 3); weight.add(4, 25); weight.add(5, 16); weight.add(6, 34); // 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.println(ans);}}// This code is contributed by pratham76 |
Python3
# Python3 implementation of the approachimport mathans, x = 0, 0 graph = [[] for i in range(100)]weight = [0]*(100) # Function that returns true if# x is a perfect squaredef isPerfectSquare(x): # Find floating point value of # square root of x sr = math.sqrt(x); # If square root is an integer return ((sr - math.floor(sr)) == 0) # Function that returns true# if n is a fibonacci numberdef isFibonacci(n): return isPerfectSquare(5 * n * n + 4) or isPerfectSquare(5 * n * n - 4) # Function to perform dfsdef dfs(node, parent): global ans # If weight of the current node # gives a fibonacci number # when x is added to it if (isFibonacci(weight[node] + x)): ans += 1 for to in graph[node]: if (to == parent): continue dfs(to, node)x = 5 # Weights of the nodeweight[1] = 4weight[2] = 5weight[3] = 3weight[4] = 25weight[5] = 16weight[6] = 34# Edges of the treegraph[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 divyesh072019. |
C#
// C# implementation of the // above approachusing System;using System.Collections; class GFG{ static int ans = 0, x; static ArrayList []graph = new ArrayList[100];static ArrayList weight = new ArrayList(); // Function that returns true if// x is a perfect squarestatic bool isPerfectSquare(double x){ // Find floating point value of // square root of x double sr = Math.Sqrt(x); // If square root is an integer return ((sr - Math.Floor(sr)) == 0);} // Function that returns true// if n is a fibonacci numberstatic bool isFibonacci(int n){ return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);} // Function to perform dfsstatic void dfs(int node, int parent){ // If weight of the current node // gives a fibonacci number // when x is added to it if (isFibonacci((int)weight[node] + x)) ans += 1; foreach (int to in graph[node]) { if (to == parent) continue; dfs(to, node); }} // Driver Codepublic static void Main(string[] args){ x = 5; for(int i = 0; i < 100; i++) { weight.Add(0); 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] = 34; // 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 // above approachvar ans = 0, x; var graph = Array.from(Array(100), ()=>Array());var weight = []; // Function that returns true if// x is a perfect squarefunction isPerfectSquare(x){ // Find floating point value of // square root of x var sr = Math.sqrt(x); // If square root is an integer return ((sr - Math.floor(sr)) == 0);} // Function that returns true// if n is a fibonacci numberfunction isFibonacci(n){ return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);} // Function to perform dfsfunction dfs(node, parent){ // If weight of the current node // gives a fibonacci number // when x is added to it if (isFibonacci(weight[node] + x)) ans += 1; for(var to of graph[node]) { if (to == parent) continue; dfs(to, node); }} // Driver Codex = 5;for(var i = 0; i < 100; i++){ weight.push(0); graph[i] = [];}// Weights of the nodeweight[1] = 4;weight[2] = 5;weight[3] = 3;weight[4] = 25;weight[5] = 16;weight[6] = 34;// Edges of the treegraph[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);</script> |
Output:
2
Time Complexity : O(n).
Auxilitary Space Complexity : O(n).
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