<p>You are given a positive integer <code>n</code>.</p>

<p>A binary string <code>x</code> is <strong>valid</strong> if all <span data-keyword="substring-nonempty">substrings</span> of <code>x</code> of length 2 contain <strong>at least</strong> one <code>&quot;1&quot;</code>.</p>

<p>Return all <strong>valid</strong> strings with length <code>n</code><strong>, </strong>in <em>any</em> order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 3</span></p>

<p><strong>Output:</strong> <span class="example-io">[&quot;010&quot;,&quot;011&quot;,&quot;101&quot;,&quot;110&quot;,&quot;111&quot;]</span></p>

<p><strong>Explanation:</strong></p>

<p>The valid strings of length 3 are: <code>&quot;010&quot;</code>, <code>&quot;011&quot;</code>, <code>&quot;101&quot;</code>, <code>&quot;110&quot;</code>, and <code>&quot;111&quot;</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">[&quot;0&quot;,&quot;1&quot;]</span></p>

<p><strong>Explanation:</strong></p>

<p>The valid strings of length 1 are: <code>&quot;0&quot;</code> and <code>&quot;1&quot;</code>.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 18</code></li>
</ul>

class Solution:
    def validStrings(self, n: int) -> List[str]:

        # Initialize DP lists
        dp = [[] for _ in range(n + 1)]
        
        # Base case for length 1
        dp[1] = ['0', '1']
        
        # Fill DP table for lengths 2 to n
        for i in range(2, n + 1):
            # For strings of length i that end with '1'
            for s in dp[i - 1]:
                dp[i].append(s + '1')
            
            # For strings of length i that end with '0'
            for s in dp[i - 1]:
                if s[-1] != '0':  # Ensure no adjacent '0'
                    dp[i].append(s + '0')
        
        return dp[n]
        
44