1980. Find Unique Binary String

Problem

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**.

  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.

  Constraints:

• n == nums.length

• 1 <= n <= 16

• nums[i].length == n

• nums[i] is either '0' or '1'.

• All the strings of nums are unique.

Solution (Java)

class Solution {
    public String findDifferentBinaryString(String[] nums) {
        Set<String> set = new HashSet<>(Arrays.asList(nums));
        int len = nums[0].length();
        StringBuilder sb = new StringBuilder();
        int i = 0;
        while (i < len) {
            sb.append(1);
            i++;
        }
        int max = Integer.parseInt(sb.toString(), 2);
        for (int num = 0; num <= max; num++) {
            String binary = Integer.toBinaryString(num);
            if (binary.length() < len) {
                sb.setLength(0);
                sb.append(binary);
                while (sb.length() < len) {
                    sb.insert(0, "0");
                }
                binary = sb.toString();
            }
            if (!set.contains(binary)) {
                return binary;
            }
        }
        return null;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).