<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 {
public:
    string findDifferentBinaryString(vector<string>& nums) {
        vector<string>allPos; 
        set<string>st; 
        for(auto it: nums) 
        st.insert(it); 

        int n = nums[0].size(); 

        for(int i=0;i<(1<<n);i++)
        {
            string s; 
            for(int k=0;k<n;k++) 
            s.push_back('0');  

            for(int j=0;j<n;j++){
                if( (i) & (1<<j) ){
                    s[j] = '1'; 
                }
            }
            allPos.push_back(s); 
        }
        
        for(auto it: allPos){
            if(st.find(it) == st.end()) return it; 
        }
        return "-1"; 
    }
};
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