<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>

#define pq priority_queue
#define um unordered_map
#define us unordered_set
#define	ll long long
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define mp make_pair

class Solution {
public:
    vector<int> gridIllumination(int n, vector<vector<int>>& lamps, vector<vector<int>>& queries) {
        vector<int> ans;
        um<int, us<int>> lamp; // x, y
        um<int, int> x, y; // Store the number of lamp on row and column that is turned on
        um<int, int> slash, rSlash; // Store the / or \ that the light is on
        // Slash is from top-right to bottom-left is i + j, top-left to bottom-right is i - j
        int len = queries.size();

        for(auto &l: lamps){
            auto in = lamp[l[0]].insert(l[1]);
            if(in.second){
                // Insert successfully
                x[l[0]]++;
                y[l[1]]++;
                slash[l[0] + l[1]]++;
                rSlash[l[0] - l[1]]++;
            }
        }
        for(int i = 0; i < len; i++){
            bool found = false;
            if(x[queries[i][0]] > 0 || y[queries[i][1]] > 0 || slash[queries[i][0] + queries[i][1]] > 0|| rSlash[queries[i][0] - queries[i][1]] > 0){
                ans.pb(1);
            }else{
                ans.pb(0);
            }

            // Turne off the lamp next to it
            vector<int> temp = queries[i]; // Start from top-left
            temp[0]--;
            temp[1]--;
            for(int idxI = 0; idxI < 3; idxI++){
                if(temp[0] + idxI < 0 && temp[0] + idxI > n) continue;
                for(int idxJ = 0; idxJ < 3; idxJ++){
                    if(temp[1] + idxJ < 0 && temp[1] + idxJ > n) continue;
                    if(lamp[temp[0] + idxI].erase(temp[1] + idxJ) == 1){
                        x[temp[0] + idxI]--;
                        y[temp[1] + idxJ]--;
                        slash[(temp[0] + idxI) + (temp[1] + idxJ)]--;
                        rSlash[(temp[0] + idxI) - (temp[1] + idxJ)]--;
                    }
                }
            }
        }

        return ans;
    }
};
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