<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:
    struct pairHash {
        size_t operator()(const pair<int, int> &x) const { return x.first ^ x.second; }
    };
    vector<int> gridIllumination(int N, vector<vector<int>>& lamps, vector<vector<int>>& queries) {
        vector<int> res;
        unordered_map<int, int> x, y, a_d, d_d;
        unordered_set<pair<int, int>, pairHash> ls;
        for (auto l : lamps) {
            auto i = l[0], j = l[1];
            if (ls.insert({ i, j }).second) 
                ++x[i], ++y[j], ++a_d[i + j], ++d_d[i - j];
        }
        for (auto q : queries) {
            auto i = q[0], j = q[1];
            if (x[i] || y[j] || a_d[i + j] || d_d[i - j]) {
            res.push_back(1);
            for (int li = i - 1; li <= i + 1; ++li)
                for (int lj = j - 1; lj <= j + 1; ++lj) {
                    if (ls.erase({li, lj})) {
                        --x[li], --y[lj], --a_d[li + lj], --d_d[li - lj];
                    }
                }
            }
            else 
                res.push_back(0);
        }
        return res;
    }
};


// The intutition to approach this answer is to store for each lamp (x,y) store it's x(for row) ,y(for col), x+y(for anti-diag), x-y(for diag) so that checking for each related (x,y) in query becomes easier.

// Thus for each (x,y) in the grid we can check the count of

// the unordered_map x --> gives the value i.e. the number of lamps in the same row responsible for lighting up the cell (x,y)
// the unordered_map y --> gives the value i.e. the number of lamps in the same col responsible for lighting up the cell (x,y)
// the unordered_map xplusy --> gives the value i.e. the number of lamps in the same anti-diag responsible for lighting up the cell (x,y).... since for same diagonal x+y remains the same.
// the unordered_map xminusy --> gives the value i.e. the number of lamps in the same diagonal responsible for lighting up the cell (x,y).... since for same diagonal x-y remains the same.
// class Solution {
// public:
// vector gridIllumination(int N, vector<vector>& lamps, vector<vector>& q) {
// vector ans;
// if(N==0)
// return ans;
// set<pair<int,int>> s;
// unordered_map<int,int> x;
// unordered_map<int,int> y;
// unordered_map<int,int> xplusy;
// unordered_map<int,int> xminusy;

//     for(int i=0;i<lamps.size();i++)
//     {
//         x[lamps[i][0]]++;      // storing count of x
//         y[lamps[i][1]]++;     // storing count of y
//         xplusy[(lamps[i][0]+lamps[i][1])]++;     // storing x+y  (to take care of anti-diag)
//         xminusy[(lamps[i][0]-lamps[i][1])]++;   // storing x+y (to take care of diagonal)
//         s.insert(make_pair(lamps[i][0],lamps[i][1]));     // storing lamp to do O(1) check later on
//     }
    
//     for(int i=0;i<q.size();i++)
//     {
//         if(x[q[i][0]]>0 || y[q[i][1]]>0 || xplusy[(q[i][0]+q[i][1])]>0 || xminusy[(q[i][0]-q[i][1])]>0)
//         {
//             ans.push_back(1);
//             int X = q[i][0];
//             int Y = q[i][1];
//             for(int j=-1;j<=1;j++)
//             {
//                 for(int k=-1;k<=1;k++)
//                 {
//                     if(X+j>=0 && X+j<N && Y+k>=0 && Y+k<N && s.find(make_pair(X+j,Y+k))!=s.end())
//                     {
//                         s.erase(make_pair(X+j,Y+k));
//                         x[X+j]--;
//                         y[Y+k]--;
//                         xplusy[(X+j+Y+k)]--;
//                         xminusy[(X+j-Y-k)]--;
//                     }
//                 }
//             }
//         }
//         else
//             ans.push_back(0);
//     }
//     return ans;
// }
// };
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