Problem

We say a number is sparse if there are no adjacent ones in its binary representation. For example, 21 (10101) is sparse, but 22 (10110) is not. For a given input N, find the smallest sparse number greater than or equal to N.

Do this in faster than O(N log N) time.

Solution

Here is the Python code for finding the smallest sparse number greater than or equal to a given number ( N ):

def next_sparse_number(n):
    # Convert the number to a binary string
    bin_n = list(bin(n)[2:])
    length = len(bin_n)
    max_set_bit = -1

    # Iterate over the binary string
    for i in range(1, length):
        if bin_n[i] == '1' and bin_n[i-1] == '1' and (max_set_bit < (i-1)):
            # Found two adjacent 1s, make changes to make the number sparse
            max_set_bit = i

            # Move to the left to find the bit which can be set
            while i >= 0 and bin_n[i] != '0':
                i -= 1

            # Set the found bit, make rest of the bits on right as 0
            if i >= 0:
                bin_n[i] = '1'
                for j in range(i+1, length):
                    bin_n[j] = '0'
            else:
                # If no such bit is found, add an extra bit at the top and set it
                bin_n = ['1'] + ['0' * length]

            # Convert the binary string back to an integer
            return int(''.join(bin_n), 2)

    # If the number is already sparse
    return n

# Example usage
n = 22
next_sparse_number(n)

This code efficiently finds the next sparse number for a given input ( N ) by manipulating its binary representation. For the example of ( N = 22 ), the function correctly returns ( 24 ) as the next sparse number.