Given a positive integer n. The problem is to 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.
Approach: If (n & (n >> 1)) == 0, then ‘n’ is a fibbinary number Else not.
C++
// C++ implementation to check whether a number// is fibbinary or not#include <bits/stdc++.h>using namespace std;// function to check whether a number// is fibbinary or notbool isFibbinaryNum(unsigned int n) { // if the number does not contain adjacent ones // then (n & (n >> 1)) operation results to 0 if ((n & (n >> 1)) == 0) return true; // not a fibbinary number return false;}// Driver program to test aboveint main() { unsigned int n = 10; if (isFibbinaryNum(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation to check whether // a number is fibbinary or notclass GFG { // function to check whether a number // is fibbinary or not static boolean isFibbinaryNum(int n) { // if the number does not contain // adjacent ones then (n & (n >> 1)) // operation results to 0 if ((n & (n >> 1)) == 0) return true; // not a fibbinary number return false; } // Driver program to test above public static void main(String[] args) { int n = 10; if (isFibbinaryNum(n) == true) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by// Smitha Dinesh Semwal |
Python3
# Python3 program to check if a number # is fibbinary number or not # function to check whether a number# is fibbinary or notdef isFibbinaryNum( n): # if the number does not contain adjacent # ones then (n & (n >> 1)) operation # results to 0 if ((n & (n >> 1)) == 0): return 1 # Not a fibbinary number return 0# Driver coden = 10if (isFibbinaryNum(n)): print("Yes")else: print("No") # This code is contributed by sunnysingh |
C#
// C# implementation to check whether // a number is fibbinary or notusing System;class GFG { // function to check whether a number // is fibbinary or not static bool isFibbinaryNum(int n) { // if the number does not contain // adjacent ones then (n & (n >> 1)) // operation results to 0 if ((n & (n >> 1)) == 0) return true; // not a fibbinary number return false; } // Driver program to test above public static void Main() { int n = 10; if (isFibbinaryNum(n) == true) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code is contributed by vt_m. |
PHP
<?php// PHP implementation to check whether// a number is fibbinary or not// function to check whether a number// is fibbinary or notfunction isFibbinaryNum($n){ // if the number does not contain // adjacent ones then (n & (n >> 1)) // operation results to 0 if (($n & ($n >> 1)) == 0) return true; // not a fibbinary number return false;}// Driver code$n = 10;if (isFibbinaryNum($n)) echo "Yes";else echo "No";// This code is contributed by mits ?> |
Javascript
<script>// JavaScript program implementation to find whether // a number is fibbinary or not // function to check whether a number // is fibbinary or not function isFibbinaryNum(n) { // if the number does not contain // adjacent ones then (n & (n >> 1)) // operation results to 0 if ((n & (n >> 1)) == 0) return true; // not a fibbinary number return false; } // Driver code let n = 10; if (isFibbinaryNum(n) == true) document.write("Yes"); else document.write("No");// This code is contributed by souravghosh0416.</script> |
Output :
Yes
Time Complexity: O(1).
Auxiliary Space: O(1).
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