Grid Illumination
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1001.Grid%20Illumination/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1001.Grid%20Illumination/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1001.Grid%20Illumination/images/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>
Array; Hash Table
class Solution {
    private int n;
    public int[] gridIllumination(int n, int[][] lamps, int[][] queries) {
        this.n = n;
        Set<Long> s = new HashSet<>();
        Map<Integer, Integer> row = new HashMap<>();
        Map<Integer, Integer> col = new HashMap<>();
        Map<Integer, Integer> diag1 = new HashMap<>();
        Map<Integer, Integer> diag2 = new HashMap<>();
        for (var lamp : lamps) {
            int i = lamp[0], j = lamp[1];
            if (s.add(f(i, j))) {
                merge(row, i, 1);
                merge(col, j, 1);
                merge(diag1, i - j, 1);
                merge(diag2, i + j, 1);
            }
        }
        int m = queries.length;
        int[] ans = new int[m];
        for (int k = 0; k < m; ++k) {
            int i = queries[k][0], j = queries[k][1];
            if (exist(row, i) || exist(col, j) || exist(diag1, i - j) || exist(diag2, i + j)) {
                ans[k] = 1;
            }
            for (int x = i - 1; x <= i + 1; ++x) {
                for (int y = j - 1; y <= j + 1; ++y) {
                    if (x < 0 || x >= n || y < 0 || y >= n || !s.contains(f(x, y))) {
                        continue;
                    }
                    s.remove(f(x, y));
                    merge(row, x, -1);
                    merge(col, y, -1);
                    merge(diag1, x - y, -1);
                    merge(diag2, x + y, -1);
                }
            }
        }
        return ans;
    }

    private void merge(Map<Integer, Integer> cnt, int x, int d) {
        if (cnt.merge(x, d, Integer::sum) == 0) {
            cnt.remove(x);
        }
    }

    private boolean exist(Map<Integer, Integer> cnt, int x) {
        return cnt.getOrDefault(x, 0) > 0;
    }

    private long f(long i, long j) {
        return i * n + j;
    }
}