<p>There is a 2D <code>grid</code> of size <code>n x n</code> where each cell of this grid has a lamp that is initially <strong>turned off</strong>.</p>

<p>You are given a 2D array of lamp positions <code>lamps</code>, where <code>lamps[i] = [row<sub>i</sub>, col<sub>i</sub>]</code> indicates that the lamp at <code>grid[row<sub>i</sub>][col<sub>i</sub>]</code> is <strong>turned on</strong>. Even if the same lamp is listed more than once, it is turned on.</p>

<p>When a lamp is turned on, it <strong>illuminates its cell</strong> and <strong>all other cells</strong> in the same <strong>row, column, or diagonal</strong>.</p>

<p>You are also given another 2D array <code>queries</code>, where <code>queries[j] = [row<sub>j</sub>, col<sub>j</sub>]</code>. For the <code>j<sup>th</sup></code> query, determine whether <code>grid[row<sub>j</sub>][col<sub>j</sub>]</code> is illuminated or not. After answering the <code>j<sup>th</sup></code> query, <strong>turn off</strong> the lamp at <code>grid[row<sub>j</sub>][col<sub>j</sub>]</code> and its <strong>8 adjacent lamps</strong> if they exist. A lamp is adjacent if its cell shares either a side or corner with <code>grid[row<sub>j</sub>][col<sub>j</sub>]</code>.</p>

<p>Return <em>an array of integers </em><code>ans</code><em>,</em><em> where </em><code>ans[j]</code><em> should be </em><code>1</code><em> if the cell in the </em><code>j<sup>th</sup></code><em> query was illuminated, or </em><code>0</code><em> if the lamp was not.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/19/illu_1.jpg" style="width: 750px; height: 209px;" />
<pre>
<strong>Input:</strong> n = 5, lamps = [[0,0],[4,4]], queries = [[1,1],[1,0]]
<strong>Output:</strong> [1,0]
<strong>Explanation:</strong> We have the initial grid with all lamps turned off. In the above picture we see the grid after turning on the lamp at grid[0][0] then turning on the lamp at grid[4][4].
The 0<sup>th</sup>&nbsp;query asks if the lamp at grid[1][1] is illuminated or not (the blue square). It is illuminated, so set ans[0] = 1. Then, we turn off all lamps in the red square.
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/19/illu_step1.jpg" style="width: 500px; height: 218px;" />
The 1<sup>st</sup>&nbsp;query asks if the lamp at grid[1][0] is illuminated or not (the blue square). It is not illuminated, so set ans[1] = 0. Then, we turn off all lamps in the red rectangle.
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/19/illu_step2.jpg" style="width: 500px; height: 219px;" />
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 5, lamps = [[0,0],[4,4]], queries = [[1,1],[1,1]]
<strong>Output:</strong> [1,1]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 5, lamps = [[0,0],[0,4]], queries = [[0,4],[0,1],[1,4]]
<strong>Output:</strong> [1,1,0]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= lamps.length &lt;= 20000</code></li>
	<li><code>0 &lt;= queries.length &lt;= 20000</code></li>
	<li><code>lamps[i].length == 2</code></li>
	<li><code>0 &lt;= row<sub>i</sub>, col<sub>i</sub> &lt; n</code></li>
	<li><code>queries[j].length == 2</code></li>
	<li><code>0 &lt;= row<sub>j</sub>, col<sub>j</sub> &lt; n</code></li>
</ul>

class Solution {
public:
    vector<int> gridIllumination(int n, vector<vector<int>>& l,
                                 vector<vector<int>>& q) {
        unordered_map<int, int> lx, ly, ld, rd;
        vector<int> a;
        vector<vector<int>> dir;
        unordered_set<long> lamp;
        for (auto& i : l) {
            long s = (long)i[0] * n + i[1];
            if (!lamp.count(s)) {
                lamp.insert(s);
                lx[i[0]]++;
                ly[i[1]]++;
                ld[i[0] - i[1]]++;
                rd[i[0] + i[1]]++;
            }
        }
        dir = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 0},
               {0, 1},   {1, -1}, {1, 0},  {1, 1}};
        for (auto& i : q) {
            int x = i[0], y = i[1];
            if (lx.count(x) || ly.count(y) || ld.count(x - y) ||
                rd.count(x + y)) {
                a.push_back(1);
            } else {
                a.push_back(0);
                continue;
            }
            for (auto& d : dir) {
                int nx = x + d[0], ny = y + d[1];
                if (nx >= 0 && nx < n && ny >= 0 && ny < n) {
                    long s = (long)nx * n + ny;
                    if (lamp.count(s)) {
                        if (lx.count(nx)) {
                            lx[nx]--;
                            if (lx[nx] < 1)
                                lx.erase(nx);
                        }
                        if (ly.count(ny)) {
                            ly[ny]--;
                            if (ly[ny] < 1)
                                ly.erase(ny);
                        }
                        if (ld.count(nx - ny)) {
                            ld[nx - ny]--;
                            if (ld[nx - ny] < 1)
                                ld.erase(nx - ny);
                        }
                        if (rd.count(nx + ny)) {
                            rd[nx + ny]--;
                            if (rd[nx + ny] < 1)
                                rd.erase(nx + ny);
                        }
                        lamp.erase(s);
                    }
                }
            }
        }
        return a;
    }
};
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