<p>Given an array of strings <code>nums</code> containing <code>n</code> <strong>unique</strong> binary strings each of length <code>n</code>, return <em>a binary string of length </em><code>n</code><em> that <strong>does not appear</strong> in </em><code>nums</code><em>. If there are multiple answers, you may return <strong>any</strong> of them</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [&quot;01&quot;,&quot;10&quot;]
<strong>Output:</strong> &quot;11&quot;
<strong>Explanation:</strong> &quot;11&quot; does not appear in nums. &quot;00&quot; would also be correct.
</pre>

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

<pre>
<strong>Input:</strong> nums = [&quot;00&quot;,&quot;01&quot;]
<strong>Output:</strong> &quot;11&quot;
<strong>Explanation:</strong> &quot;11&quot; does not appear in nums. &quot;10&quot; would also be correct.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [&quot;111&quot;,&quot;011&quot;,&quot;001&quot;]
<strong>Output:</strong> &quot;101&quot;
<strong>Explanation:</strong> &quot;101&quot; does not appear in nums. &quot;000&quot;, &quot;010&quot;, &quot;100&quot;, and &quot;110&quot; would also be correct.
</pre>

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

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 16</code></li>
	<li><code>nums[i].length == n</code></li>
	<li><code>nums[i] </code>is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
	<li>All the strings of <code>nums</code> are <strong>unique</strong>.</li>
</ul>

class Solution:
    def findDifferentBinaryString(self, nums: List[str]) -> str:
        n = len(nums)
        nums_set = set(nums)
        
        def generate_binary_strings(length):
            if length == 0:
                return ['']
            else:
                prev_strings = generate_binary_strings(length - 1)
                result = []
                for s in prev_strings:
                    result.append(s + '0')
                    result.append(s + '1')
                return result

        for candidate in generate_binary_strings(n):
            if candidate not in nums_set:
                return candidate

# Example usage:
# sol = Solution()
# print(sol.findDifferentBinaryString(["01","10"]))  # Output: "11"
# print(sol.findDifferentBinaryString(["00","01"]))  # Output: "11"
# print(sol.findDifferentBinaryString(["111","011","001"]))  # Output: "101"
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