<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 thereIsLamp(self, lmps, a, b):
        tmp = str(a) + "," + str(b)
        if tmp in lmps:
            x = lmps[tmp]
            del lmps[tmp]
            return x
        return 0

    def turnOffLamp(self, lmps, rows, cols, diag1, diag2, n, a, b):
        x = self.thereIsLamp(lmps, a, b)
        if x:
            d1 = b-a
            d2 = b-(n-1-a)
            rows[a] -= x
            cols[b] -= x
            diag1[d1] -= x
            diag2[d2] -= x

    def gridIllumination(self, n: int, lamps: List[List[int]], queries: List[List[int]]) -> List[int]:
        # turn on all lamps
        lmps = dict()
        for a,b in lamps:
            tmp = str(a) + "," + str(b)
            if tmp not in lmps:
                lmps[tmp] = 0
            lmps[tmp] += 1

        # 1. in hash-maps: count how many lamps per row, column, diagonal, other-diagonal
        rows, cols, diag1, diag2 = dict(), dict(), dict(), dict()
        for a,b in lamps:
            if a not in rows: rows[a] = 0
            rows[a] += 1

            if b not in cols: cols[b] = 0
            cols[b] += 1

            d1 = b-a
            if d1 not in diag1: diag1[d1] = 0
            diag1[d1] += 1

            d2 = b-(n-1-a)
            if d2 not in diag2: diag2[d2] = 0
            diag2[d2] += 1
        #print(rows, cols, diag1, diag2)
        # 2. for each query:
            # a) if lamp in this row, column, diagonal, other-diagonal, say False, otherwise True
            # b) for each of the 8 neighbor cells and the query-cell: do binary-search if there are turned on lamps, if there is, subtract 1 from the according row, column, diagonal and other-diagonal, and label those lamps as turned off 
        res = []
        for a,b in queries:
            d1 = b-a
            d2 = b-(n-1-a)
            v1 = (a in rows) and rows[a]
            v2 = (b in cols) and cols[b]
            v3 = (d1 in diag1) and diag1[d1]
            v4 = (d2 in diag2) and diag2[d2]
            if v1 or v2 or v3 or v4:
                res.append(1)
            else:
                res.append(0)

            for i in range(-1, 2):
                for j in range(-1, 2):
                    self.turnOffLamp(lmps, rows, cols, diag1, diag2, n, a+i, b+j)

        return res
31700