<p>You are given a <strong>0-indexed</strong> 2D matrix <code>grid</code> of size <code>n x n</code>, where <code>(r, c)</code> represents:</p>

<ul>
	<li>A cell containing a thief if <code>grid[r][c] = 1</code></li>
	<li>An empty cell if <code>grid[r][c] = 0</code></li>
</ul>

<p>You are initially positioned at cell <code>(0, 0)</code>. In one move, you can move to any adjacent cell in the grid, including cells containing thieves.</p>

<p>The <strong>safeness factor</strong> of a path on the grid is defined as the <strong>minimum</strong> manhattan distance from any cell in the path to any thief in the grid.</p>

<p>Return <em>the <strong>maximum safeness factor</strong> of all paths leading to cell </em><code>(n - 1, n - 1)</code><em>.</em></p>

<p>An <strong>adjacent</strong> cell of cell <code>(r, c)</code>, is one of the cells <code>(r, c + 1)</code>, <code>(r, c - 1)</code>, <code>(r + 1, c)</code> and <code>(r - 1, c)</code> if it exists.</p>

<p>The <strong>Manhattan distance</strong> between two cells <code>(a, b)</code> and <code>(x, y)</code> is equal to <code>|a - x| + |b - y|</code>, where <code>|val|</code> denotes the absolute value of val.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example1.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0],[0,0,0],[0,0,1]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All paths from (0, 0) to (n - 1, n - 1) go through the thieves in cells (0, 0) and (n - 1, n - 1).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example2.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,1],[0,0,0],[0,0,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The path depicted in the picture above has a safeness factor of 2 since:
- The closest cell of the path to the thief at cell (0, 2) is cell (0, 0). The distance between them is | 0 - 0 | + | 0 - 2 | = 2.
It can be shown that there are no other paths with a higher safeness factor.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example3.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0,1],[0,0,0,0],[0,0,0,0],[1,0,0,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The path depicted in the picture above has a safeness factor of 2 since:
- The closest cell of the path to the thief at cell (0, 3) is cell (1, 2). The distance between them is | 0 - 1 | + | 3 - 2 | = 2.
- The closest cell of the path to the thief at cell (3, 0) is cell (3, 2). The distance between them is | 3 - 3 | + | 0 - 2 | = 2.
It can be shown that there are no other paths with a higher safeness factor.
</pre>

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

<ul>
	<li><code>1 &lt;= grid.length == n &lt;= 400</code></li>
	<li><code>grid[i].length == n</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There is at least one thief in the <code>grid</code>.</li>
</ul>

class Solution {
public:
    bool valid(int row,int col,int n,int m){
        if(row<0 ||row>=n || col<0 || col>=m){
            return false;
        }
        return true;
    }
    bool check(int mid,int n,int m,vector<vector<int>>&dis){
        if(dis[0][0]<mid)return false;
        queue<pair<int,int>>q;
        q.push({0,0});
        int delrow[4]={0,1,0,-1};
        int delcol[4]={1,0,-1,0};
        vector<vector<int>>vis(n,vector<int>(m,0));
        vis[0][0]=1;
        int flag=0;
        while(!q.empty()){
            auto it=q.front();
            int row=it.first;
            int col=it.second;
            q.pop();
            if(row==n-1 && col==n-1){
                flag=1;
                break;
            }
            for(int i=0;i<4;i++){
                int nrow=row+delrow[i];
                int ncol=col+delcol[i];
                if(valid(nrow,ncol,n,m) && vis[nrow][ncol]==0 && dis[nrow][ncol]>=mid){
                    vis[nrow][ncol]=1;
                    q.push({nrow,ncol});
                }
            }
        }
        if(flag==1){
            return true;
        }
        return false;
    }
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        queue<pair<pair<int,int>,int>>q;
        int n=grid.size();
        int m=grid[0].size();
        map<pair<int,int>,int>mp2;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==1){
                    q.push({{i,j},0});
                    mp2[{i,j}]=1;
                }
            }
        }
        vector<vector<int>>dis(n,vector<int>(m,1e9+7));
        int delrow[4]={0,1,0,-1};
        int delcol[4]={1,0,-1,0};
        while(!q.empty()){
            auto it=q.front();
            int row=it.first.first;
            int col=it.first.second;
            int curr_dis=it.second;
            q.pop();
            for(int i=0;i<4;i++){
                int nrow=row+delrow[i];
                int ncol=col+delcol[i];
                if(valid(nrow,ncol,n,m) && mp2[{nrow,ncol}]==0){
                  if(curr_dis+1<dis[nrow][ncol]){
                     dis[nrow][ncol]=curr_dis+1;
                     q.push({{nrow,ncol},dis[nrow][ncol]});
                   }
                }
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(dis[i][j]==1e9+7){
                    dis[i][j]=0;
                }
                // cout<<dis[i][j]<<" ";
            }
            // cout<<endl;
        }
        // cout<<endl;
        int lo=0;
        int hi=2*n;
        int ans=0;
        while(lo<=hi){
            int mid=lo+(hi-lo)/2;
            if(check(mid,n,m,dis)){
                lo=mid+1;
                ans=mid;
            }
            else{
                hi=mid-1;
            }
        }
        return ans;
    }
};
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