{"id":9285,"date":"2021-12-30T03:07:41","date_gmt":"2021-12-30T11:07:41","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=9285"},"modified":"2021-12-30T03:10:19","modified_gmt":"2021-12-30T11:10:19","slug":"leetcode-1926-nearest-exit-from-entrance-in-maze","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/searching\/leetcode-1926-nearest-exit-from-entrance-in-maze\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1926. Nearest Exit from Entrance in Maze"},"content":{"rendered":"\n

You are given an m x n<\/code> matrix maze<\/code> (0-indexed<\/strong>) with empty cells (represented as '.'<\/code>) and walls (represented as '+'<\/code>). You are also given the entrance<\/code> of the maze, where entrance = [entrancerow<\/sub>, entrancecol<\/sub>]<\/code> denotes the row and column of the cell you are initially standing at.<\/p>\n\n\n\n

In one step, you can move one cell up<\/strong>, down<\/strong>, left<\/strong>, or right<\/strong>. You cannot step into a cell with a wall, and you cannot step outside the maze. Your goal is to find the nearest exit<\/strong> from the entrance<\/code>. An exit<\/strong> is defined as an empty cell<\/strong> that is at the border<\/strong> of the maze<\/code>. The entrance<\/code> does not count<\/strong> as an exit.<\/p>\n\n\n\n

Return the number of steps<\/strong> in the shortest path from the <\/em>entrance<\/code> to the nearest exit, or <\/em>-1<\/code> if no such path exists<\/em>.<\/p>\n\n\n\n

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

\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> maze = [[\"+\",\"+\",\".\",\"+\"],[\".\",\".\",\".\",\"+\"],[\"+\",\"+\",\"+\",\".\"]], entrance = [1,2]\nOutput:<\/strong> 1\nExplanation:<\/strong> There are 3 exits in this maze at [1,0], [0,2], and [2,3].\nInitially, you are at the entrance cell [1,2].\n- You can reach [1,0] by moving 2 steps left.\n- You can reach [0,2] by moving 1 step up.\nIt is impossible to reach [2,3] from the entrance.\nThus, the nearest exit is [0,2], which is 1 step away.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> maze = [[\"+\",\"+\",\"+\"],[\".\",\".\",\".\"],[\"+\",\"+\",\"+\"]], entrance = [1,0]\nOutput:<\/strong> 2\nExplanation:<\/strong> There is 1 exit in this maze at [1,2].\n[1,0] does not count as an exit since it is the entrance cell.\nInitially, you are at the entrance cell [1,0].\n- You can reach [1,2] by moving 2 steps right.\nThus, the nearest exit is [1,2], which is 2 steps away.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> maze = [[\".\",\"+\"]], entrance = [0,0]\nOutput:<\/strong> -1\nExplanation:<\/strong> There are no exits in this maze.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • maze.length == m<\/code><\/li>
  • maze[i].length == n<\/code><\/li>
  • 1 <= m, n <= 100<\/code><\/li>
  • maze[i][j]<\/code> is either '.'<\/code> or '+'<\/code>.<\/li>
  • entrance.length == 2<\/code><\/li>
  • 0 <= entrancerow<\/sub> < m<\/code><\/li>
  • 0 <= entrancecol<\/sub> < n<\/code><\/li>
  • entrance<\/code> will always be an empty cell.<\/li><\/ul>\n\n\n\n
    Solution: BFS<\/strong><\/h2>\n\n\n\n
    Use BFS to find the shortest path. We can re-use the board for visited array.<\/p>\n\n\n\n
    Time complexity: O(m*n)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int nearestExit(vector>& maze, vector& entrance) {    \n    const int m = maze.size();\n    const int n = maze[0].size();\n    const vector dirs{0, -1, 0, 1, 0};\n    queue> q;\n    q.emplace(entrance[1], entrance[0]);    \n    for (int steps = 0; !q.empty(); ++steps) {      \n      for (int s = q.size(); s; --s) {      \n        const auto [x, y] = q.front(); q.pop();        \n        if (x == 0 || x == n - 1 || y == 0 || y == m - 1)\n          if (x != entrance[1] || y != entrance[0])\n            return steps;\n        for (int i = 0; i < 4; ++i) {\n          const int tx = x + dirs[i];\n          const int ty = y + dirs[i + 1];\n          if (tx < 0 || tx >= n || ty < 0 || ty >= m || maze[ty][tx] != '.') continue;\n          maze[ty][tx] = '*';\n          q.emplace(tx, ty);\n        }\n      }      \n    }\n    return -1;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    You are given an m x n matrix maze (0-indexed) with empty cells (represented as ‘.’) and walls (represented as ‘+’). You are also given the entrance of the maze, where entrance = [entrancerow,…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[44],"tags":[34,87],"class_list":["post-9285","post","type-post","status-publish","format-standard","hentry","category-searching","tag-bfs","tag-shortest-path","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9285","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=9285"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9285\/revisions"}],"predecessor-version":[{"id":9287,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9285\/revisions\/9287"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=9285"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=9285"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=9285"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}