Fibbinary Numbers (No consecutive 1s in binary)
Given N, check if the number is a Fibbinary number or not. Fibbinary numbers are integers whose binary representation includes 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 of doing 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 fibbinary number.
Some of the first few Fibbinary 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 1sbool 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 functionint 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 notclass 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 1sdef 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 coden = 10if (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 notusing 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 1sfunction 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 ?> |
Javascript
<script> // javascript 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 var 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 var n = 10; if (checkFibinnary(n) == true) document.write("YES"); else document.write("NO");// This code contributed by Rajput-Ji</script> |
Output:
YES
Time Complexity: O(logN), as we are using a loop to traverse logN times, we are decrementing by floor division of 2 (as right shifting a number by 1 is equivalent to floor division by 2) in each iteration therefore the loop iterates logN times.
Auxiliary Space: O(1), as we are not using any extra space.
Fibbinary Numbers (No consecutive 1s in binary) – O(1) Approach
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!
