<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]:
        ans = list()
        hor = dict()
        ver = dict()
        diagL = dict() # stores left diagonal line
        diagR = dict() # stores right diagonal line
        hashTable = dict() # stores lamps in hash table
        lamps = [list(i) for i in set(map(tuple, lamps))] #removes duplicate lamps

        for i in lamps:
            hashTable[(i[0], i[1])] = True
            if i[0] in hor:
                hor[i[0]] += 1
            else:
                hor[i[0]] = 1

            if i[1] in ver:
                ver[i[1]] += 1
            else:
                ver[i[1]] = 1

            if i[0] - i[1] in diagL:
                diagL[i[0] - i[1]] += 1
            else:
                diagL[i[0] - i[1]] = 1

            if i[0] + i[1] in diagR:
                diagR[i[0] + i[1]] += 1
            else:
                diagR[i[0] + i[1]] = 1

        for i in queries:
            row, col = i
            if row in hor or col in ver or row - col in diagL or row + col in diagR:
                ans.append(1)
                for j in range(9):
                    x, y = int(row - 1 + j / 3), col - 1 + j % 3
                    checkPos = (x, y)
                    if x < 0 or x >= n or y < 0 or y >= n:
                        continue

                    if x in hor and checkPos in hashTable:
                        hor[x] -= 1
                        if hor[x] == 0:
                            hor.pop(x)

                    if y in ver and checkPos in hashTable:
                        ver[y] -= 1
                        if ver[y] == 0:
                            ver.pop(y)

                    if x - y in diagL and checkPos in hashTable:
                        diagL[x - y] -= 1
                        if diagL[x - y] == 0:
                            diagL.pop(x - y)

                    if x + y in diagR and checkPos in hashTable:
                        diagR[x + y] -= 1
                        if diagR[x + y] == 0:
                            diagR.pop(x + y)
                    if checkPos in hashTable:
                        hashTable.pop(checkPos)
            else:
                ans.append(0)
        return ans


        
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