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

/**
 * @param {string[]} nums
 * @return {string}
 */
var findDifferentBinaryString = function(nums) {
    var str = '';
    for (var i = 0; i <= nums.length; i++) {
        str = i.toString(2);
        str = '0'.repeat(nums.length - str.length) + str;
        if (!nums.includes(str)) {
            return str;
        }
    }
};

Explain:

Since array nums only contains n numbers, if we got n + 1 numbers, there mush have at lease one number not in that array.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).