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

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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `// function to check whether a number ` `// is fibbinary or not ` `bool` `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 above ` `int` `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 not ` `class` `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 fibinnary number or not  ` ` `  `# function to check whether a number ` `# is fibbinary or not ` `def` `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 code ` `n ``=` `10` ` `  `if` `(isFibbinaryNum(n)): ` `    ``print``(``"Yes"``) ` `else``: ` `    ``print``(``"No"``) ` `     `  `# This code is contributed by sunnysingh  `

## C#

 `// C# implementation to check whether  ` `// a number is fibbinary or not ` `using` `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

 `> 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  ` `?> `

Output :

```Yes
```

Time Complexity: O(1).

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