<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 {
    string res;
    string curr;
    unordered_map<string, int> m;
    
public:
    string findDifferentBinaryString(vector<string>& nums) {
        int n = nums.size();
        string curr = "";

        for(int i=0; i<nums.size(); i++){
            m[nums[i]] = 1;
        }

        dfs(0, n, curr);

        return res;

        
    }

    void dfs(int i, int n, string curr){
        if(i == n){
            if(m.find(curr)!=m.end()){
                return;
            }
            else{
                res = curr;
                return;
            }            
        }

        curr += "0";
        dfs(i+1, n, curr);
        curr.pop_back();

        curr += '1';        
        dfs(i+1, n, curr);
        curr.pop_back();
        return;
    }
};
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