3

Intuition

Code

Python3

def validstrings(self, n: int) -> list[str]:
 
    result = []
 
    def helper(s):
        '''
        recursively build string s, append a zero only if previous char is not a zero
 
        append a one regardless
        '''
        if len(s) == n:
            result.append(s)
            return
 
        helper(s + '1')
        if s[-1] == '1':
            helper(s + '0')
 
    helper('0')  # for binary strings starting with zeros
    helper('1')  # for binary strings starting with ones
 
    return result

Big O Analysis

• Runtime

  The runtime complexity here is $O(2^N)$ since there can be $2^N$ combinations.

• Memory

  The memory usage is $O(N)$ since we use a list to store the generated strings.

— A

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