Largest number with binary representation is m 1's and m-1 0's

Given n, find the greatest number which is strictly not more than n and whose binary representation consists of m consecutive ones, then m-1 consecutive zeros and nothing else

Examples:

Input : n = 7 
Output : 6 
Explanation: 6's binary representation is 110,
and 7's is 111, so 6 consists of 2 consecutive 
1's and then 1 consecutive 0.

Input : 130
Output : 120 
Explanation: 28 and 120 are the only numbers <=120,
28 is 11100 consists of 3 consecutive 1's and then 
2 consecutive 0's. 120 is 1111000 consists of 4 
consecutive 1's and then 3 consecutive 0's. So 120
is the greatest of number<=120 which meets the
given condition. 

A naive approach will be to traverse from 1 to N and check for every binary representation which consists of m consecutive 1's and m-1 consecutive 0's and store the largest of them which meets the given condition. 

An efficient approach is to observe a pattern of numbers,
[1(1), 6(110), 28(11100), 120(1111000), 496(111110000), ....] 
To get the formula for the numbers which satisfies the conditions we take 120 as an example- 
120 is represented as 1111000 which has m = 4 1's and m = 3 0's. Converting 1111000 to decimal we get: 
2^3+2^4+2^5+2^6 which can be represented as (2^m-1 + 2^m+ 2^m+1 + ... 2^m+2, 2^2*m) 
2^3*(1+2+2^2+2^3) which can be represented as (2^(m-1)*(1+2+2^2+2^3+..2^(m-1)) 
2^3*(2^4-1) which can be represented as [2^(m-1) * (2^m -1)].

So all the numbers that meet the given condition can be represented as 

[2^(m-1) * (2^m -1)]

We can iterate till the number does not exceeds N and print the largest of all possible elements. A closer observation will shows that at m = 33 it will exceed the 10^18 mark , so we are calculating the number in unit's time as log(32) is near to constant which is required in calculating the pow . 
So, the overall complexity will be O(1). 

6

Time Complexity: O(1)
Auxiliary Space: O(1)

Efficient approach:

This problem can be solved with at most two comparisons, by using the following logic:

1. For a number with n bits, the answer with m set bits followed by m - 1 bits can have a total of n bits, ie, m =  (n + 1)/2 (as m + m - 1 = n).

2. The result can then be computed as 2 ^{(m + m - 1)} - 2 ^{(m - 1)}.

3. However, if the n - bit number is in the range [2 ^{(m + m - 2)}, 2 ^{(m + m - 1)} - 2 ^{(m - 1)} - 1],  then m must be adjusted to m - 1.

Below is the implementation of the above approach:

6

Time Complexity: O(1)
Auxiliary Space: O(1)