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1980. Find Unique Binary String

Description

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.

Solutions

  • class Solution {
        public String findDifferentBinaryString(String[] nums) {
            int mask = 0;
            for (var x : nums) {
                int cnt = 0;
                for (int i = 0; i < x.length(); ++i) {
                    if (x.charAt(i) == '1') {
                        ++cnt;
                    }
                }
                mask |= 1 << cnt;
            }
            for (int i = 0;; ++i) {
                if ((mask >> i & 1) == 0) {
                    return "1".repeat(i) + "0".repeat(nums.length - i);
                }
            }
        }
    }
    
  • class Solution {
    public:
        string findDifferentBinaryString(vector<string>& nums) {
            int mask = 0;
            for (auto& x : nums) {
                int cnt = count(x.begin(), x.end(), '1');
                mask |= 1 << cnt;
            }
            for (int i = 0;; ++i) {
                if (mask >> i & 1 ^ 1) {
                    return string(i, '1') + string(nums.size() - i, '0');
                }
            }
        }
    };
    
  • class Solution:
        def findDifferentBinaryString(self, nums: List[str]) -> str:
            mask = 0
            for x in nums:
                mask |= 1 << x.count("1")
            n = len(nums)
            for i in range(n + 1):
                if mask >> i & 1 ^ 1:
                    return "1" * i + "0" * (n - i)
    
    
  • func findDifferentBinaryString(nums []string) string {
    	mask := 0
    	for _, x := range nums {
    		mask |= 1 << strings.Count(x, "1")
    	}
    	for i := 0; ; i++ {
    		if mask>>i&1 == 0 {
    			return strings.Repeat("1", i) + strings.Repeat("0", len(nums)-i)
    		}
    	}
    }
    
  • public class Solution {
        public string FindDifferentBinaryString(string[] nums) {
            int mask = 0;
            foreach (var x in nums) {
                int cnt = x.Count(c => c == '1');
                mask |= 1 << cnt;
            }
            int i = 0;
            while ((mask >> i & 1) == 1) {
                i++;
            }
            return string.Format("{0}{1}", new string('1', i), new string('0', nums.Length - i));
        }
    }
    
    

All Problems

All Solutions