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
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