<p>You are given an integer <code>n</code>. You have an <code>n x n</code> binary grid <code>grid</code> with all values initially <code>1</code>&#39;s except for some indices given in the array <code>mines</code>. The <code>i<sup>th</sup></code> element of the array <code>mines</code> is defined as <code>mines[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> where <code>grid[x<sub>i</sub>][y<sub>i</sub>] == 0</code>.</p>

<p>Return <em>the order of the largest <strong>axis-aligned</strong> plus sign of </em>1<em>&#39;s contained in </em><code>grid</code>. If there is none, return <code>0</code>.</p>

<p>An <strong>axis-aligned plus sign</strong> of <code>1</code>&#39;s of order <code>k</code> has some center <code>grid[r][c] == 1</code> along with four arms of length <code>k - 1</code> going up, down, left, and right, and made of <code>1</code>&#39;s. Note that there could be <code>0</code>&#39;s or <code>1</code>&#39;s beyond the arms of the plus sign, only the relevant area of the plus sign is checked for <code>1</code>&#39;s.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/13/plus1-grid.jpg" style="width: 404px; height: 405px;" />
<pre>
<strong>Input:</strong> n = 5, mines = [[4,2]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> In the above grid, the largest plus sign can only be of order 2. One of them is shown.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/13/plus2-grid.jpg" style="width: 84px; height: 85px;" />
<pre>
<strong>Input:</strong> n = 1, mines = [[0,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no plus sign, so return 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>1 &lt;= mines.length &lt;= 5000</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt; n</code></li>
	<li>All the pairs <code>(x<sub>i</sub>, y<sub>i</sub>)</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    int orderOfLargestPlusSign(int n, vector<vector<int>>& nums) {
        map<pair<int,int>,int>mp;
        for(auto &i:nums)
            mp[{i[0],i[1]}]++;
        vector<vector<pair<int,int>>>h(n,vector<pair<int,int>>(n,{0,0}));
        vector<vector<pair<int,int>>>v(n,vector<pair<int,int>>(n,{0,0}));
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(i==0){
                    if(mp.find({i,j})==mp.end())
                        v[i][j].first=1;
                    if(mp.find({n-i-1,j})==mp.end())
                        v[n-i-1][j].second=1;
                }
                else{
                    if(mp.find({i,j})==mp.end())
                        v[i][j].first=v[i-1][j].first+1;
                    if(mp.find({n-i-1,j})==mp.end())
                        v[n-i-1][j].second=v[n-i][j].second+1;
                }
                if(j==0){
                    if(mp.find({i,j})==mp.end())
                        h[i][j].first=1;
                    if(mp.find({i,n-j-1})==mp.end())
                        h[i][n-j-1].second=1;
                }
                else{
                    if(mp.find({i,j})==mp.end())
                        h[i][j].first=h[i][j-1].first+1;
                    if(mp.find({i,n-j-1})==mp.end())
                        h[i][n-j-1].second=h[i][n-j].second+1;
                }
            }
        }
        int ans=0;
        if(mp.size()!=n*n)ans++;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(mp.find({i,j})!=mp.end())continue;
                int x=min(h[i][j].second-1,h[i][j].first-1);
                x=max(0,x);
                int y=min(v[i][j].second-1,v[i][j].first-1);
                y=max(0,y);
                int t=min(x,y);
                if(t==0)continue;
                ans=max(ans,t+1);
            }
        }
        // for(int i=0;i<n;i++){
        //     for(int j=0;j<n;j++){
        //         if(mp.find({i,j})==mp.end())cout<<"1 ";
        //         else cout<<"0 ";
        //     }
        //     cout<<endl;
        // }
        // cout<<endl;
        // for(auto &i:v){
        //     for(auto &k:i){
        //         cout<<k.second<<" ";
        //     }
        //     cout<<endl;
        // }
        // cout<<"_____\n";
        // for(auto &i:h){
        //     for(auto &k:i){
        //         cout<<k.second<<" ";
        //     }
        //     cout<<endl;
        // }
        return ans;
    }
};
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