Problem Description

You are given a positive integer n.

A binary string x is valid if all substrings of x of length 2 contain at least one "1".

Return all valid strings with length n, in any order.

Intuition

To solve this problem, we use a Depth-First Search (DFS) approach to explore every possible binary string combination of length n.

The key is to ensure that every substring of length 2 within the generated binary string contains at least one "1". This requirement implies:

• If the current bit is 0, the previous bit must be 1 to ensure the substring formed by this and the previous bit contains at least one 1.
• If the current bit is 1, we can freely choose the previous bit as the condition is met inherently.

Starting from an empty string, we iteratively build each position with the above conditions using recursion until the string reaches length n. This ensures we explore all valid permutations. Each time we reach a string of length n, we add it to the list of valid strings. This form of backtracking ensures we cover all possible options, checking validity at each step.

Learn more about Backtracking patterns.

Solution Approach

The solution uses Depth-First Search (DFS) to generate all valid binary strings of length n. Let's delve into the specifics:

1. Recursive Function (dfs): We define a recursive helper function dfs(i) where i is the current index in the string being formed. The goal is to fill up the string one bit at a time until it reaches length n.

2. Base Case: If i equals n, we have a complete binary string of the desired length. At this point, we join the list t (holding the current string) into a string and add it to the list ans.

3. Exploring Each Bit:

   • We consider both 0 and 1 at each position i.
   • If the current bit j is 0, it can only be placed if it follows a 1. This check is vital since a 0 needs to satisfy the valid substring condition together with its predecessor.
   • If the bit j is 1, it is always valid to place, as any substring ending with 1 meets the criteria.

4. Backtracking: After placing a bit and recursively filling the next position, we remove the bit (using pop) and backtrack to try other possibilities if needed. This backtracking is crucial to explore different combinations efficiently.

5. Data Structures:

   • A list t is used to build and store the current configuration of the binary string during each recursive call.
   • A list ans stores all valid string configurations of length n.

Here is the code for the approach:

class Solution:
    def validStrings(self, n: int) -> List[str]:
        def dfs(i: int):
            if i >= n:
                ans.append("".join(t))
                return
            for j in range(2):
                if (j == 0 and (i == 0 or t[i - 1] == "1")) or j == 1:
                    t.append(str(j))
                    dfs(i + 1)
                    t.pop()

        ans = []
        t = []
        dfs(0)
        return ans

This algorithm efficiently generates all possible valid combinations by recursively constructing each possibility and ensuring it adheres to the condition that every pair of consecutive bits in the string contains at least one 1.

Example Walkthrough

Let's walk through the solution approach with a small example where n = 3.

The task is to generate all valid binary strings of length 3, ensuring every substring of length 2 contains at least one "1".

Step-by-Step Walkthrough using DFS Approach:

1. Initialization:

   • We start with an empty list t to build our string and an empty list ans to store all valid strings.

2. Recursive Exploration:

   • We begin by calling dfs(0) to start filling the string from index 0.

3. Index 0 Exploration:

   • Option 1: Place "1" at index 0.
     • Update t to ['1'].
     • Call dfs(1) to move to the next index.
   • Option 2: Skip "0" at index 0 since the rule requires the preceding bit to be "1".

4. Index 1 Exploration with t = ['1']:

   • Option 1: Place "0" at index 1 (valid after "1").
     • Update t to ['1', '0'].
     • Call dfs(2).
   • Option 2: Place "1" at index 1.
     • Update t to ['1', '1'].
     • Call dfs(2).

5. Index 2 Exploration:

   • With t = ['1', '0']:
     • Only Option: Place "1" at index 2 to make t = ['1', '0', '1'].
     • Call dfs(3).
     • Result: Add "101" to ans.
   • With t = ['1', '1']:
     • Option 1: Place "0" at index 2 for a valid string, resulting in t = ['1', '1', '0'].
     • Option 2: Place "1" at index 2 for another valid string, resulting in t = ['1', '1', '1'].
     • Call dfs(3) in both scenarios.
     • Results: Add "110" and "111" to ans.

6. Conclusion:

   • After backtracking and exploring all possibilities, the ans list contains: ["101", "110", "111"].

Thus, the valid strings of length 3 that meet the criteria are 101, 110, and 111.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n * 2^n). This arises because there are 2^n possible combinations of binary strings of length n. For each valid string, the code constructs the string which takes O(n) time, resulting in O(n * 2^n) overall time complexity.

The space complexity is O(n), which is primarily used for the recursive call stack and the temporary list t. The t list holds characters up to a maximum of n at any recursive depth, and the recursive depth itself can also reach n, resulting in a space complexity of O(n).

Learn more about how to find time and space complexity quickly.