LeetCode in Kotlin

1980. Find Unique Binary String

Medium

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

class Solution {
    fun findDifferentBinaryString(nums: Array<String>): String {
        val set: Set<String> = HashSet(listOf(*nums))
        val len = nums[0].length
        val sb = StringBuilder()
        var i = 0
        while (i < len) {
            sb.append(1)
            i++
        }
        val max = sb.toString().toInt(2)
        for (num in 0..max) {
            var 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 ""
    }
}

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