{"id":9359,"date":"2021-12-31T15:38:17","date_gmt":"2021-12-31T23:38:17","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=9359"},"modified":"2021-12-31T15:40:06","modified_gmt":"2021-12-31T23:40:06","slug":"leetcode-1980-find-unique-binary-string","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/bit\/leetcode-1980-find-unique-binary-string\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1980. Find Unique Binary String"},"content":{"rendered":"\n

Given an array of strings nums<\/code> containing n<\/code> unique<\/strong> binary strings each of length n<\/code>, return a binary string of length <\/em>n<\/code> that does not appear<\/strong> in <\/em>nums<\/code>. If there are multiple answers, you may return any<\/strong> of them<\/em>.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> nums = [\"01\",\"10\"]\nOutput:<\/strong> \"11\"\nExplanation:<\/strong> \"11\" does not appear in nums. \"00\" would also be correct.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [\"00\",\"01\"]\nOutput:<\/strong> \"11\"\nExplanation:<\/strong> \"11\" does not appear in nums. \"10\" would also be correct.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [\"111\",\"011\",\"001\"]\nOutput:<\/strong> \"101\"\nExplanation:<\/strong> \"101\" does not appear in nums. \"000\", \"010\", \"100\", and \"110\" would also be correct.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • n == nums.length<\/code><\/li>
  • 1 <= n <= 16<\/code><\/li>
  • nums[i].length == n<\/code><\/li>
  • nums[i] <\/code>is either '0'<\/code> or '1'<\/code>.<\/li>
  • All the strings of nums<\/code> are unique<\/strong>.<\/li><\/ul>\n\n\n\n
    Solution 1: Hashtable<\/strong><\/h2>\n\n\n\n
    We can use bitset to convert between integer and binary string.<\/p>\n\n\n\n
    Time complexity: O(n2<\/sup>)
    Space complexity: O(n2<\/sup>)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  string findDifferentBinaryString(vector& nums) {\n    const int n = nums.size();\n    unordered_set seen(1 << n);\n    for (const string& num : nums)\n      seen.insert(bitset<16>(num).to_ulong());\n    for (int i = 0; i < 1 << n; ++i)\n      if (!seen.count(i)) return bitset<16>(i).to_string().substr(16 - n);\n    return \"\";\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    Solution 2: One bit a time<\/strong><\/h2>\n\n\n\n
    Let ans[i] = ‘1’ – nums[i][i], s.t. ans is at least one bit different from any strings.<\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  string findDifferentBinaryString(vector& nums) {    \n    const int n = nums.size();\n    string ans(n, '0');\n    for (int i = 0; i < n; ++i)\n      ans[i] = '1' - nums[i][i] + '0';\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    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.…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[126],"tags":[39,177,4],"class_list":["post-9359","post","type-post","status-publish","format-standard","hentry","category-bit","tag-binary","tag-medium","tag-string","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9359","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=9359"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9359\/revisions"}],"predecessor-version":[{"id":9362,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9359\/revisions\/9362"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=9359"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=9359"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=9359"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}