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

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 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# 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:

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).

Article Tags :

DSA

Recursion

Tree

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