{"id":7983,"date":"2021-01-17T12:39:55","date_gmt":"2021-01-17T20:39:55","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=7983"},"modified":"2021-01-17T12:40:47","modified_gmt":"2021-01-17T20:40:47","slug":"leetcode-1725-number-of-rectangles-that-can-form-the-largest-square","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/geometry\/leetcode-1725-number-of-rectangles-that-can-form-the-largest-square\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1725. Number Of Rectangles That Can Form The Largest Square"},"content":{"rendered":"\n

You are given an array rectangles<\/code> where rectangles[i] = [li<\/sub>, wi<\/sub>]<\/code> represents the ith<\/sup><\/code> rectangle of length li<\/sub><\/code> and width wi<\/sub><\/code>.<\/p>\n\n\n\n

You can cut the ith<\/sup><\/code> rectangle to form a square with a side length of k<\/code> if both k <= li<\/sub><\/code> and k <= wi<\/sub><\/code>. For example, if you have a rectangle [4,6]<\/code>, you can cut it to get a square with a side length of at most 4<\/code>.<\/p>\n\n\n\n

Let maxLen<\/code> be the side length of the largest<\/strong> square you can obtain from any of the given rectangles.<\/p>\n\n\n\n

Return the number<\/strong> of rectangles that can make a square with a side length of <\/em>maxLen<\/code>.<\/p>\n\n\n\n

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

Input:<\/strong> rectangles = [[5,8],[3,9],[5,12],[16,5]]\nOutput:<\/strong> 3\nExplanation:<\/strong> The largest squares you can get from each rectangle are of lengths [5,3,5,5].\nThe largest possible square is of length 5, and you can get it out of 3 rectangles.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> rectangles = [[2,3],[3,7],[4,3],[3,7]]\nOutput:<\/strong> 3\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= rectangles.length <= 1000<\/code><\/li>
  • rectangles[i].length == 2<\/code><\/li>
  • 1 <= li<\/sub>, wi<\/sub> <= 109<\/sup><\/code><\/li>
  • li<\/sub> != wi<\/sub><\/code><\/li><\/ul>\n\n\n\n
    Solution: Running Max of Shortest Edge<\/strong><\/h2>\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  int countGoodRectangles(vector>& rectangles) {\n    int cur = 0;\n    int ans = 0;\n    for (const auto& r : rectangles) {\n      if (min(r[0], r[1]) > cur) {\n        cur = min(r[0], r[1]);\n        ans = 0;\n      }\n      if (min(r[0], r[1]) == cur) ++ans;\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    You are given an array rectangles where rectangles[i] = [li, wi] represents the ith rectangle of length li and width wi. You can cut the ith rectangle to form a square with a side length of k if…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[127],"tags":[222,353],"class_list":["post-7983","post","type-post","status-publish","format-standard","hentry","category-geometry","tag-easy","tag-gemoetry","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7983","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=7983"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7983\/revisions"}],"predecessor-version":[{"id":7985,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7983\/revisions\/7985"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=7983"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=7983"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=7983"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}