Given N, check if the number is Fibbinary Number or not. Fibbinary numbers are integers whose binary representation contains no consecutive ones.

Examples :

Input : 10 Output : YES Explanation: 1010 is the binary representation of 10 which does not contains any consecutive 1's. Input : 11 Output : NO Explanation: 1011 is the binary representation of 11, which contains consecutive 1's

The idea to do this is to right shift the number, till n!=0. For every binary representation of 1, check if the last bit found was 1 or not. Get the last bit of binary representation of the integer by doing a (n&1). If the last bit of the binary representation is 1 and the previous bit before doing a right shift was also one, we encounter consecutive 1’s. So we come to the conclusion that it is not a fibonnary number.

Some of the first few Fibonnary numbers are:

0, 2, 4, 8, 10, 16, 18, 20.......

## CPP

// CPP program to check if a number // is fibinnary number or not #include <iostream> using namespace std; // function to check if binary // representation of an integer // has consecutive 1s bool checkFibinnary(int n) { // stores the previous last bit // initially as 0 int prev_last = 0; while (n) { // if current last bit and // previous last bit is 1 if ((n & 1) && prev_last) return false; // stores the last bit prev_last = n & 1; // right shift the number n >>= 1; } return true; } // Driver code to check above function int main() { int n = 10; if (checkFibinnary(n)) cout << "YES"; else cout << "NO"; return 0; }

## Java

// Java program to check if a number // is fibinnary number or not class GFG { // function to check if binary // representation of an integer // has consecutive 1s static boolean checkFibinnary(int n) { // stores the previous last bit // initially as 0 int prev_last = 0; while (n != 0) { // if current last bit and // previous last bit is 1 if ((n & 1) != 0 && prev_last != 0) return false; // stores the last bit prev_last = n & 1; // right shift the number n >>= 1; } return true; } // Driver code to check above function public static void main(String[] args) { int n = 10; if (checkFibinnary(n) == true) System.out.println("YES"); else System.out.println("NO"); } } // This code is contributed by // Smitha Dinesh Semwal

## Python3

# Python 3 program to check if a # number is fibinnary number or # not # function to check if binary # representation of an integer # has consecutive 1s def checkFibinnary(n): # stores the previous last bit # initially as 0 prev_last = 0 while (n): # if current last bit and # previous last bit is 1 if ((n & 1) and prev_last): return False # stores the last bit prev_last = n & 1 # right shift the number n >>= 1 return True # Driver code n = 10 if (checkFibinnary(n)): print("YES") else: print("NO") # This code is contributed by Smitha Dinesh Semwal

## C#

// C# program to check if a number // is fibinnary number or not using System; class GFG { // function to check if binary // representation of an integer // has consecutive 1s static bool checkFibinnary(int n) { // stores the previous last bit // initially as 0 int prev_last = 0; while (n != 0) { // if current last bit and // previous last bit is 1 if ((n & 1) != 0 && prev_last != 0) return false; // stores the last bit prev_last = n & 1; // right shift the number n >>= 1; } return true; } // Driver code to check above function public static void Main() { int n = 10; if (checkFibinnary(n) == true) Console.WriteLine("YES"); else Console.WriteLine("NO"); } } // This code is contributed by vt_m.

## PHP

<?php // PHP program to check if a number // is fibinnary number or not // function to check if binary // representation of an integer // has consecutive 1s function checkFibinnary($n) { // stores the previous last bit // initially as 0 $prev_last = 0; while ($n) { // if current last bit and // previous last bit is 1 if (($n & 1) && $prev_last) return false; // stores the last bit $prev_last = $n & 1; // right shift the number $n >>= 1; } return true; } // Driver code $n = 10; if (checkFibinnary($n)) echo "YES"; else echo "NO"; // This code is contributed by mits ?>

**Output :**

YES

**Time Complexity:** O( log(n) )

Fibbinary Numbers (No consecutive 1s in binary) – O(1) Approach

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