{"id":4896,"date":"2019-02-24T10:18:24","date_gmt":"2019-02-24T18:18:24","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=4896"},"modified":"2019-02-24T10:26:47","modified_gmt":"2019-02-24T18:26:47","slug":"leetcode-1001-grid-illumination","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-1001-grid-illumination\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1001. Grid Illumination"},"content":{"rendered":"\n

On a N x N<\/code> grid of cells, each cell (x, y)<\/code> with 0 <= x < N<\/code> and 0 <= y < N<\/code> has a lamp.<\/p>\n\n\n\n

Initially, some number of lamps are on.  lamps[i]<\/code> tells us the location of the i<\/code>-th lamp that is on.  Each lamp that is on illuminates every square on its x-axis, y-axis, and both diagonals (similar to a Queen in chess).<\/p>\n\n\n\n

For the i-th query queries[i] = (x, y)<\/code>, the answer to the query is 1 if the cell (x, y) is illuminated, else 0.<\/p>\n\n\n\n

After each query (x, y)<\/code> [in the order given by queries<\/code>], we turn off any lamps that are at cell (x, y)<\/code> or are adjacent 8-directionally (ie., share a corner or edge with cell (x, y)<\/code>.)<\/p>\n\n\n\n

Return an array of answers.  Each value answer[i]<\/code> should be equal to the answer of the i<\/code>-th query queries[i]<\/code>.<\/p>\n\n\n\n

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

Input: <\/strong>N = 5, lamps = [[0,0],[4,4]], queries = [[1,1],[1,0]]\nOutput: <\/strong>[1,0]\nExplanation: <\/strong>\nBefore performing the first query we have both lamps [0,0] and [4,4] on.\nThe grid representing which cells are lit looks like this, where [0,0] is the top left corner, and [4,4] is the bottom right corner:\n1 1 1 1 1\n1 1 0 0 1\n1 0 1 0 1\n1 0 0 1 1\n1 1 1 1 1\nThen the query at [1, 1] returns 1 because the cell is lit.  After this query, the lamp at [0, 0] turns off, and the grid now looks like this:\n1 0 0 0 1\n0 1 0 0 1\n0 0 1 0 1\n0 0 0 1 1\n1 1 1 1 1\nBefore performing the second query we have only the lamp [4,4] on.  Now the query at [1,0] returns 0, because the cell is no longer lit.\n<\/pre>\n\n\n\n
Note:<\/strong><\/p>\n\n\n\n
  1. 1 <= N <= 10^9<\/code><\/li>
  2. 0 <= lamps.length <= 20000<\/code><\/li>
  3. 0 <= queries.length <= 20000<\/code><\/li>
  4. lamps[i].length == queries[i].length == 2<\/code><\/li><\/ol>\n\n\n\n
    Solution: HashTable<\/strong><\/h2>\n\n\n\n
    use lx, ly, lp, lq to track the # of lamps that covers each row, col, diagonal, antidiagonal<\/p>\n\n\n\n
    Time complexity: O(|L| + |Q|)
    Space complexity: O(|L|)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua, running time: 460 ms, 82.2 MB\nclass Solution {\npublic:\n  vector gridIllumination(int N, vector>& lamps, vector>& queries) {\n    unordered_set s;\n    unordered_map lx, ly, lp, lq;\n    for (const auto& lamp : lamps) {\n      const int x = lamp[0];\n      const int y = lamp[1];\n      s.insert(static_cast(x) << 32 | y);\n      ++lx[x];\n      ++ly[y];\n      ++lp[x + y];\n      ++lq[x - y];\n    }\n    vector ans;\n    for (const auto& query : queries) {\n      const int x = query[0];\n      const int y = query[1];      \n      if (lx.count(x) || ly.count(y) || lp.count(x + y) || lq.count(x - y)) {\n        ans.push_back(1);       \n        for (int tx = x - 1; tx <= x + 1; ++tx)\n          for (int ty = y - 1; ty <= y + 1; ++ty) {\n            if (tx < 0 || tx >= N || ty < 0 || ty >= N) continue;\n            const long key = static_cast(tx) << 32 | ty;\n            if (!s.count(key)) continue;\n            s.erase(key);\n            if (--lx[tx] == 0) lx.erase(tx);\n            if (--ly[ty] == 0) ly.erase(ty);\n            if (--lp[tx + ty] == 0) lp.erase(tx + ty);\n            if (--lq[tx - ty] == 0) lq.erase(tx - ty);\n          }\n      } else {\n        ans.push_back(0);\n      }\n    }\n    return ans;\n  }\n};\n<\/pre>\n\n<\/div>
    C++ v2<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua, running time: 444 ms, 82.2 MB\nclass Solution {\npublic:\n  vector gridIllumination(int N, vector>& lamps, vector>& queries) {\n    unordered_set s;\n    unordered_map lx, ly, lp, lq;\n    for (const auto& lamp : lamps) {\n      const int x = lamp[0];\n      const int y = lamp[1];\n      s.insert(static_cast(x) << 32 | y);\n      ++lx[x];\n      ++ly[y];\n      ++lp[x + y];\n      ++lq[x - y];\n    }\n    vector ans;\n    auto dec = [](unordered_map& m, int key) {\n      auto it = m.find(key);\n      if (it->second == 1)\n        m.erase(it);\n      else\n        --it->second;\n    };\n    for (const auto& query : queries) {\n      const int x = query[0];\n      const int y = query[1];      \n      if (lx.count(x) || ly.count(y) || lp.count(x + y) || lq.count(x - y)) {\n        ans.push_back(1);       \n        for (int tx = x - 1; tx <= x + 1; ++tx)\n          for (int ty = y - 1; ty <= y + 1; ++ty) {\n            if (tx < 0 || tx >= N || ty < 0 || ty >= N) continue;\n            const long key = static_cast(tx) << 32 | ty;\n            auto it = s.find(key);\n            if (it == s.end()) continue;\n            s.erase(it);\n            dec(lx, tx);\n            dec(ly, ty);\n            dec(lp, tx + ty);\n            dec(lq, tx - ty);            \n          }\n      } else {\n        ans.push_back(0);\n      }\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
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