<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:
    def gridIllumination(self, n: int, lamps: List[List[int]], queries: List[List[int]]) -> List[int]:
        # Use dictionaries to count how many lamps are illuminating a particular row, column, diagonal, and anti-diagonal
        row_count = {}
        col_count = {}
        diag_count = {}
        anti_diag_count = {}
        
        # Set to track the positions of lamps that are currently on
        lamp_set = set()

        # Initialize the lamp state by populating the dictionaries and the lamp_set
        for r, c in lamps:
            if (r, c) in lamp_set:
                continue  # Skip if the lamp is already in the set (avoid duplicates)
            
            # Mark the lamp as on in the set
            lamp_set.add((r, c))

            # Increment row, column, diagonal, and anti-diagonal counters
            row_count[r] = row_count.get(r, 0) + 1
            col_count[c] = col_count.get(c, 0) + 1
            diag_count[r - c] = diag_count.get(r - c, 0) + 1
            anti_diag_count[r + c] = anti_diag_count.get(r + c, 0) + 1

        # Direction vectors for adjacent cells (and the cell itself)
        directions = [(0, 0), (-1, 0), (1, 0), (0, -1), (0, 1), (-1, -1), (-1, 1), (1, -1), (1, 1)]

        # List to store the answer for each query
        result = []

        # Process each query
        for qr, qc in queries:
            # Check if the query cell (qr, qc) is illuminated
            if (row_count.get(qr, 0) > 0 or
                col_count.get(qc, 0) > 0 or
                diag_count.get(qr - qc, 0) > 0 or
                anti_diag_count.get(qr + qc, 0) > 0):
                result.append(1)  # The cell is illuminated
            else:
                result.append(0)  # The cell is not illuminated

            # Turn off the lamp at the query cell and its adjacent cells (if they exist)
            for dr, dc in directions:
                nr, nc = qr + dr, qc + dc  # Calculate the new row and column for adjacent cells

                if (nr, nc) in lamp_set:
                    # Remove the lamp from the set
                    lamp_set.remove((nr, nc))

                    # Decrement the row, column, diagonal, and anti-diagonal counters
                    row_count[nr] -= 1
                    if row_count[nr] == 0:
                        del row_count[nr]  # Remove entry if no lamps remain in this row
                    
                    col_count[nc] -= 1
                    if col_count[nc] == 0:
                        del col_count[nc]  # Remove entry if no lamps remain in this column
                    
                    diag_count[nr - nc] -= 1
                    if diag_count[nr - nc] == 0:
                        del diag_count[nr - nc]  # Remove entry if no lamps remain in this diagonal
                    
                    anti_diag_count[nr + nc] -= 1
                    if anti_diag_count[nr + nc] == 0:
                        del anti_diag_count[nr + nc]  # Remove entry if no lamps remain in this anti-diagonal

        return result  # Return the list of results for each query

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