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

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Example 1:

Input: nums = ["01","10"]
Output: "11"
Explanation: "11" does not appear in nums. "00" would also be correct.


Example 2:

Input: nums = ["00","01"]
Output: "11"
Explanation: "11" does not appear in nums. "10" would also be correct.


Example 3:

Input: nums = ["111","011","001"]
Output: "101"
Explanation: "101" does not appear in nums. "000", "010", "100", and "110" would also be correct.


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Constraints:


	n == nums.length
	1 &lt;= n &lt;= 16
	nums[i].length == n
	nums[i] is either '0' or '1'.
	All the strings of nums are unique.


class Solution(object):
    def findDifferentBinaryString(self, nums):
        ans='';
        for i,num in enumerate(nums):
            ans+= '1' if(num[i]=='0') else '0'       #ternary if else
			#ans+= str(1- int(num[i]));              # Alternate: cast to string & 1-x to flip
        return ans;
class Solution {
    public String findDifferentBinaryString(String[] nums) {
        StringBuilder ans= new StringBuilder();                  
        for(int i=0; i<nums.length; i++)  
            ans.append(nums[i].charAt(i) == '0' ? '1' : '0');              // Using ternary operator
        return ans.toString();
    }
}
class Solution {
public:
    string findDifferentBinaryString(vector<string>& nums) {
        unordered_set<int> s;
        for (auto num : nums) s.insert(stoi(num, 0, 2));
        int res = 0;
        while (++res) {
            if (!s.count(res)) return bitset<16>(res).to_string().substr(16-nums.size());
        }
        return "";
    }
};
var findDifferentBinaryString = function(nums) {
    return nums.map((s, i) => s[i] == 1 ? '0' : '1').join('');
};
Find Unique Binary String