<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:
    vector<int> gridIllumination(int n, vector<vector<int>>& lamps, vector<vector<int>>& queries) {
        set<pair<int,int>> st;
        map<int,int> row,col,diag1,diag2;
        

        for(int i =0; i < lamps.size();i++){
            int r = lamps[i][0],c = lamps[i][1];
             if (st.insert({r, c}).second) {  // Only add if it's not already in the set
                row[r]++;
                col[c]++;
                diag1[r + c]++;
                diag2[n-1+ c-r]++;
            
        }}
        vector<int> ans;
        

        for(int j = 0 ; j < queries.size();j++){
            int r = queries[j][0] , c = queries[j][1];
            if(row[r] >0 || col[c] > 0 || diag1[r + c] > 0 || diag2[n-1+c-r] >0){
                ans.push_back(1);
            }
            else{
                ans.push_back(0);
            }
            remove(n,c,r,st,row,col,diag1,diag2);
        }

        return ans;
    }
    void remove(int n,int col,int row, set<pair<int,int>> &st,map<int,int> &row1,map<int,int> &col1,map<int,int> &diag1,map<int,int> &diag2){
        for(int i = -1; i <= 1;i++){
            for(int j = -1; j <= 1;j++){
                int nrow = row + i;
                int ncol = col + j;
                if(nrow >= 0 && ncol >= 0 && nrow  < n && ncol < n && st.find({nrow,ncol}) != st.end()){
                    st.erase({nrow,ncol});
                    int r = nrow,c = ncol;

                    if(--row1[r] == 0) row1.erase(r);
                    if(--col1[c] == 0) col1.erase(c);
                    if(--diag1[r+c] == 0) diag1.erase(r+c);
                    if(--diag2[n-1+c-r] == 0) diag2.erase(n-1+c-r);
                }
            }
        }
    }
    // bool isill(int n, int col, int row, set<pair<int,int>> &st) {
    
    // for (int c = 0; c < n; c++) {
    //         if(st.find({row, c}) != st.end()) return true;
    //     }
        
       
    //     for (int r = 0; r < n; r++) {
    //         if(st.find({r, col}) != st.end()) return true;
    //     }
        
    //     for (int r = row, c = col; r >= 0 && c >= 0; r--, c--) {
    //         if(st.find({r, c}) != st.end()) return true;
    //     }
    //     for (int r = row, c = col; r < n && c < n; r++, c++) {
    //         if(st.find({r, c}) != st.end()) return true;
    //     }
        
    //     for (int r = row, c = col; r >= 0 && c < n; r--, c++) {
    //         if(st.find({r, c}) != st.end()) return true;
    //     }
    //     for (int r = row, c = col; r < n && c >= 0; r++, c--) {
    //         if(st.find({r, c}) != st.end()) return true;
    //     }

    //     return false;
    // }

};
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