Fibbinary Numbers (No consecutive 1s in binary) - O(1) Approach Last Updated : 15 Jun, 2022 Comments Improve Suggest changes Like Article Like Report 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 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 fibbinary 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 <?php // PHP implementation to check 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 $n = 10; if (isFibbinaryNum($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). 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