<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>>& lamps, vector<vector<int>>& queries) {
        
        unordered_map<int,int> row, col, d1, d2;
        unordered_map<long long, bool> cell; // Use a 1D representation instead of pair<int, int>
        
        // Mark the lamps and populate the maps
        for(auto& x : lamps){
            int i = x[0], j = x[1];
            long long pos = (long long)i * n + j; // Convert (i, j) to a unique 1D position
            if(cell.find(pos) == cell.end() || !cell[pos]){
                row[i]++;
                col[j]++;
                d1[i + j]++;
                d2[i - j + n - 1]++;
                cell[pos] = true;
            }
        }
        
        vector<vector<int>> dir = {{0,0},{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,-1},{1,-1},{-1,1}};
        vector<int> result(queries.size(), 0);
        
        for(int k = 0; k < queries.size(); k++){
            int i = queries[k][0], j = queries[k][1];
            long long pos = (long long)i * n + j;
            
            if(row.find(i) != row.end() || col.find(j) != col.end() || d1.find(i + j) != d1.end() || d2.find(i - j + n - 1) != d2.end()){
                result[k] = 1;
            }
            

            for(auto& y : dir){
                int ni = i + y[0], nj = j + y[1];
                long long newPos = (long long)ni * n + nj; 
                
                if(ni >= 0 && ni < n && nj >= 0 && nj < n && cell.find(newPos) != cell.end() && cell[newPos]){
                
                    row[ni]--;
                    if(row[ni] == 0) row.erase(ni);
                    
                    col[nj]--;
                    if(col[nj] == 0) col.erase(nj);
                    
                    d1[ni + nj]--;
                    if(d1[ni + nj] == 0) d1.erase(ni + nj);
                    
                    d2[ni - nj + n - 1]--;
                    if(d2[ni - nj + n - 1] == 0) d2.erase(ni - nj + n - 1);
                    
                    cell[newPos] = false; 
                }
            }
        }
        
        return result;
    }
};

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