Check if GCD of all Composite Numbers in an array divisible by K is a Fibonacci Number or not
Last Updated :
27 Apr, 2021
Given array arr[] consisting of N non-negative integers, and an integer K, the task is to check if GCD of all composite numbers in the array which are divisible by K is a Fibonacci Number or not. IF found to be true, print "Yes". Otherwise, print "No".
Examples:
Input: arr[] = {13, 55, 1331, 7, 13, 11, 44, 77, 144, 89}, K = 11
Output: No
Explanation: Composite Numbers from the array which are divisible by 11 are {55, 1331, 11, 44, 77}. GCD of these elements is equal to 11, which is not a Fibonacci Number.
Input: arr[] = {34, 2, 4, 8, 5, 7, 11}, K = 2
Output:Yes
Explanation: Composite Numbers from the array which are divisible by 2 are {34, 2, 4, 8}. GCD of these elements is equal to 2, which is not a Fibonacci Number.
Approach: Follow the steps below to solve the problem:
- Create a function isComposite() to check if a number is a composite number or not.
- Create another function isFibonacci() to check if a number is Fibonacci number or not.
- Initialize a vector of integers, say compositeset, and another integer variable gcd to store gcd of the composite numbers from the array which are divisible by K.
- Traverse the array arr[].
- For every element arr[i], check if it is composite and divisible by K or not. If found to be true, insert it into the vector compositeset
- Calculate GCD of all the elements in the vector compositeset and store it in the variable gcd.
- Check if gcd is a Fibonacci Number or not.
- If found to be true, then print "Yes". Otherwise, print "No".
Below is the implementation of the above approach:
C++
// C++ Program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if a
// number is composite or not
bool isComposite(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return false;
// Check if the number is
// divisible by 2 or 3 or not
if (n % 2 == 0 || n % 3 == 0)
return true;
// Check if n is a multiple of
// any other prime number
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return true;
return false;
}
// Function to check if a number
// is a Perfect Square or not
bool isPerfectSquare(int x)
{
int s = sqrt(x);
return (s * s == x);
}
// Function to check if a number
// is a Fibonacci number or not
bool isFibonacci(int n)
{
// If 5*n^2 + 4 or 5*n^2 - 4 or
// both are perfect square
return isPerfectSquare(5 * n * n + 4)
|| isPerfectSquare(5 * n * n - 4);
}
// Function to check if GCD of composite
// numbers from the array a[] which are
// divisible by k is a Fibonacci number or not
void ifgcdFibonacci(int a[], int n, int k)
{
vector<int> compositeset;
// Traverse the array
for (int i = 0; i < n; i++) {
// If array element is composite
// and divisible by k
if (isComposite(a[i]) && a[i] % k == 0) {
compositeset.push_back(a[i]);
}
}
int gcd = compositeset[0];
// Calculate GCD of all elements in compositeset
for (int i = 1; i < compositeset.size(); i++) {
gcd = __gcd(gcd, compositeset[i]);
if (gcd == 1) {
break;
}
}
// If GCD is Fibonacci
if (isFibonacci(gcd)) {
cout << "Yes";
return;
}
cout << "No";
return;
}
// Driver Code
int main()
{
int arr[] = { 34, 2, 4, 8, 5, 7, 11 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 2;
ifgcdFibonacci(arr, n, k);
return 0;
}
// Java Program for the above approach
import java.util.*;
class GFG{
// Function to check if a
// number is composite or not
static boolean isComposite(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return false;
// Check if the number is
// divisible by 2 or 3 or not
if (n % 2 == 0 || n % 3 == 0)
return true;
// Check if n is a multiple of
// any other prime number
for(int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return true;
return false;
}
// Function to check if a number
// is a Perfect Square or not
static boolean isPerfectSquare(int x)
{
int s = (int)Math.sqrt(x);
return (s * s == x);
}
// Function to check if a number
// is a Fibonacci number or not
static boolean isFibonacci(int n)
{
// If 5*n^2 + 4 or 5*n^2 - 4 or
// both are perfect square
return isPerfectSquare(5 * n * n + 4) ||
isPerfectSquare(5 * n * n - 4);
}
// Function to check if GCD of composite
// numbers from the array a[] which are
// divisible by k is a Fibonacci number or not
static void ifgcdFibonacci(int a[], int n, int k)
{
Vector<Integer> compositeset = new Vector<>();
// Traverse the array
for(int i = 0; i < n; i++)
{
// If array element is composite
// and divisible by k
if (isComposite(a[i]) && a[i] % k == 0)
{
compositeset.add(a[i]);
}
}
int gcd = compositeset.get(0);
// Calculate GCD of all elements in compositeset
for(int i = 1; i < compositeset.size(); i++)
{
gcd = __gcd(gcd, compositeset.get(i));
if (gcd == 1)
{
break;
}
}
// If GCD is Fibonacci
if (isFibonacci(gcd))
{
System.out.print("Yes");
return;
}
System.out.print("No");
return;
}
// Recursive function to return gcd of a and b
static int __gcd(int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 34, 2, 4, 8, 5, 7, 11 };
int n = arr.length;
int k = 2;
ifgcdFibonacci(arr, n, k);
}
}
// This code is contributed by Amit Katiyar
# Python3 program for the above approach
import math
# Function to check if a
# number is composite or not
def isComposite(n):
# Corner cases
if n <= 1:
return False
if n <= 3:
return False
# Check if the number is
# divisible by 2 or 3 or not
if n % 2 == 0 or n % 3 == 0:
return True
# Check if n is a multiple of
# any other prime number
i = 5
while i * i <= n:
if ((n % i == 0 ) or
(n % (i + 2) == 0)):
return True
i += 6
return False
# Function to check if a number
# is a Perfect Square or not
def isPerfectSquare(x):
s = int(math.sqrt(x))
return (s * s == x)
# Function to check if a number
# is a Fibonacci number or not
def isFibonacci(n):
# If 5*n^2 + 4 or 5*n^2 - 4 or
# both are perfect square
return (isPerfectSquare(5 * n * n + 4) or
isPerfectSquare(5 * n * n - 4))
# Function to check if GCD of composite
# numbers from the array a[] which are
# divisible by k is a Fibonacci number or not
def ifgcdFibonacci(a, n, k):
compositeset = []
# Traverse the array
for i in range(n):
# If array element is composite
# and divisible by k
if (isComposite(a[i]) and a[i] % k == 0):
compositeset.append(a[i])
gcd = compositeset[0]
# Calculate GCD of all elements in compositeset
for i in range(1, len(compositeset), 1):
gcd = math.gcd(gcd, compositeset[i])
if gcd == 1:
break
# If GCD is Fibonacci
if (isFibonacci(gcd)):
print("Yes")
return
print("No")
return
# Driver Code
if __name__ == "__main__" :
arr = [ 34, 2, 4, 8, 5, 7, 11 ]
n = len(arr)
k = 2
ifgcdFibonacci(arr, n, k)
# This code is contributed by jana_sayantan
// C# Program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to check if a
// number is composite or not
static bool isComposite(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return false;
// Check if the number is
// divisible by 2 or 3 or not
if (n % 2 == 0 || n % 3 == 0)
return true;
// Check if n is a multiple of
// any other prime number
for(int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return true;
return false;
}
// Function to check if a number
// is a Perfect Square or not
static bool isPerfectSquare(int x)
{
int s = (int)Math.Sqrt(x);
return (s * s == x);
}
// Function to check if a number
// is a Fibonacci number or not
static bool isFibonacci(int n)
{
// If 5*n^2 + 4 or 5*n^2 - 4 or
// both are perfect square
return isPerfectSquare(5 * n * n + 4) ||
isPerfectSquare(5 * n * n - 4);
}
// Function to check if GCD of composite
// numbers from the array []a which are
// divisible by k is a Fibonacci number or not
static void ifgcdFibonacci(int []a, int n, int k)
{
List<int> compositeset = new List<int>();
// Traverse the array
for(int i = 0; i < n; i++)
{
// If array element is composite
// and divisible by k
if (isComposite(a[i]) && a[i] % k == 0)
{
compositeset.Add(a[i]);
}
}
int gcd = compositeset[0];
// Calculate GCD of all elements in compositeset
for(int i = 1; i < compositeset.Count; i++)
{
gcd = __gcd(gcd, compositeset[i]);
if (gcd == 1)
{
break;
}
}
// If GCD is Fibonacci
if (isFibonacci(gcd))
{
Console.Write("Yes");
return;
}
Console.Write("No");
return;
}
// Recursive function to return gcd of a and b
static int __gcd(int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 34, 2, 4, 8, 5, 7, 11 };
int n = arr.Length;
int k = 2;
ifgcdFibonacci(arr, n, k);
}
}
// This code is contributed by 29AjayKumar
<script>
// Javascript Program for the above approach
function __gcd(a, b) {
if (!b) {
return a;
}
return __gcd(b, a % b);
}
// Function to check if a
// number is composite or not
function isComposite(n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return false;
// Check if the number is
// divisible by 2 or 3 or not
if (n % 2 == 0 || n % 3 == 0)
return true;
var i;
// Check if n is a multiple of
// any other prime number
for(i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return true;
return false;
}
// Function to check if a number
// is a Perfect Square or not
function isPerfectSquare(x)
{
var s = Math.sqrt(x);
return (s * s == x);
}
// Function to check if a number
// is a Fibonacci number or not
function isFibonacci(n)
{
// If 5*n^2 + 4 or 5*n^2 - 4 or
// both are perfect square
return isPerfectSquare(5 * n * n + 4)
|| isPerfectSquare(5 * n * n - 4);
}
// Function to check if GCD of composite
// numbers from the array a[] which are
// divisible by k is a Fibonacci number or not
function ifgcdFibonacci(a, n, k)
{
var compositeset = [];
var i;
// Traverse the array
for (i = 0; i < n; i++) {
// If array element is composite
// and divisible by k
if (isComposite(a[i]) && a[i] % k == 0) {
compositeset.push(a[i]);
}
}
var gcd = compositeset[0];
// Calculate GCD of all elements in compositeset
for (i = 1; i < compositeset.length; i++) {
gcd = __gcd(gcd, compositeset[i]);
if (gcd == 1) {
break;
}
}
// If GCD is Fibonacci
if (isFibonacci(gcd)) {
document.write("Yes");
return;
}
document.write("No");
return;
}
// Driver Code
var arr = [34, 2, 4, 8, 5, 7, 11];
var n = arr.length;
var k = 2;
ifgcdFibonacci(arr, n, k);
</script>
Time Complexity: O(N*log(N)), where N is the size of the array
Auxiliary Space: O(N)
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