1 to n bit numbers with no consecutive 1s in binary representation.

Given a number n, our task is to find all 1 to n bit numbers with no consecutive 1s in their binary representation. 
Examples: 
 

Input : n = 4
Output : 1 2 4 5 8 9 10
These are numbers with 1 to 4
bits and no consecutive ones in
binary representation.

Input : n = 3
Output : 1 2 4 5

We add bits one by one and recursively print numbers. For every last bit, we have two choices. 
 

 if last digit in sol is 0 then
 we can insert 0 or 1 and recur. 
 else if last digit is 1 then
 we can insert 0 only and recur.

We will use recursion- 
 

1. We make a solution vector sol and insert first bit 1 in it which will be the first number.
2. Now we check whether length of solution vector is less than or equal to n or not.
3. If it is so then we calculate the decimal number and store it into a map as it store numbers in sorted order.
4. Now we will have two conditions- 
   
   • if last digit in sol is 0 the we can insert 0 or 1 and recur.
   • else if last digit is 1 then we can insert 0 only and recur.

numberWithNoConsecutiveOnes(n, sol)
{
if sol.size() <= n
 
 // calculate decimal and store it
 if last element of sol is 1
 insert 0 in sol 
 numberWithNoConsecutiveOnes(n, sol)
 else
 insert 1 in sol
 numberWithNoConsecutiveOnes(n, sol)

 // because we have to insert zero 
 // also in place of 1
 sol.pop_back();
 insert 0 in sol
 numberWithNoConsecutiveOnes(n, sol)
 }

Output : 
 

1 2 4 5 8 9 10

Time Complexity : O(nlogn)

Auxiliary Space: O(n)

Related Post : 
Count number of binary strings without consecutive 1’s