{"id":5900,"date":"2019-12-01T13:06:00","date_gmt":"2019-12-01T21:06:00","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=5900"},"modified":"2019-12-01T13:11:52","modified_gmt":"2019-12-01T21:11:52","slug":"leetcode-1274-number-of-ships-in-a-rectangle","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/divide-and-conquer\/leetcode-1274-number-of-ships-in-a-rectangle\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1274. Number of Ships in a Rectangle"},"content":{"rendered":"\n

(This problem is an interactive problem<\/strong>.)<\/em><\/p>\n\n\n\n

On the sea represented by a cartesian plane, each ship is located at an integer point, and each integer point may contain at most 1 ship.<\/p>\n\n\n\n

You have a function Sea.hasShips(topRight, bottomLeft)<\/code> which takes two points as arguments and returns true<\/code> if and only if there is at least one ship in the rectangle represented by the two points, including on the boundary.<\/p>\n\n\n\n

Given two points, which are the top right and bottom left corners of a rectangle, return the number of ships present in that rectangle.  It is guaranteed that there are at most 10 ships<\/strong> in that rectangle.<\/p>\n\n\n\n

Submissions making more than 400 calls<\/strong> to hasShips<\/code> will be judged Wrong Answer<\/em>.  Also, any solutions that attempt to circumvent the judge will result in disqualification.<\/p>\n\n\n\n

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

\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> \nships = [[1,1],[2,2],[3,3],[5,5]], topRight = [4,4], bottomLeft = [0,0]\nOutput:<\/strong> 3\nExplanation:<\/strong> From [0,0] to [4,4] we can count 3 ships within the range.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • On the input ships<\/code> is only given to initialize the map internally. You must solve this problem “blindfolded”. In other words, you must find the answer using the given hasShips<\/code> API, without knowing the ships<\/code> position.<\/li>
  • 0 <= bottomLeft[0] <= topRight[0] <= 1000<\/code><\/li>
  • 0 <= bottomLeft[1] <= topRight[1] <= 1000<\/code><\/li><\/ul>\n\n\n\n
    Solution: Divide and Conquer<\/strong><\/h2>\n\n\n\n
    If the current rectangle contains ships, subdivide it into 4 smaller ones until 
    1) no ships contained
    2) the current rectangle is a single point (e.g. topRight == bottomRight)<\/p>\n\n\n\n
    Time complexity: O(logn)
    Space complexity: O(logn)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  \/\/ Modify the interface to pass sea as a reference.\n  int countShips(Sea& sea, vector topRight, vector bottomLeft) {\n    int x1 = bottomLeft[0], y1 = bottomLeft[1];\n    int x2 = topRight[0], y2 = topRight[1];    \n    if (x1 > x2 || y1 > y2 || !sea.hasShips(topRight, bottomLeft))\n      return 0;\n    if (x1 == x2 && y1 == y2)\n      return 1;\n    int xm = x1 + (x2 - x1) \/ 2;\n    int ym = y1 + (y2 - y1) \/ 2;\n    return countShips(sea, {xm, ym}, {x1, y1}) + countShips(sea, {xm, y2}, {x1, ym + 1})\n         + countShips(sea, {x2, ym}, {xm + 1, y1}) + countShips(sea, {x2, y2}, {xm + 1, ym + 1});\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    (This problem is an interactive problem.) On the sea represented by a cartesian plane, each ship is located at an integer point, and each integer point…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[43],"tags":[313,217,253,341],"class_list":["post-5900","post","type-post","status-publish","format-standard","hentry","category-divide-and-conquer","tag-divide-and-conquer","tag-hard","tag-interactive","tag-quad-tree","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5900","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=5900"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5900\/revisions"}],"predecessor-version":[{"id":5902,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5900\/revisions\/5902"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=5900"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=5900"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=5900"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}