<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:
    void relax(int& r,int& c,vector<vector<int>>& safeD,vector<vector<int>>& grid,queue<pair<int,int>>& q)
    {
        int i,j,n=grid.size();
        q.push({r,c});
        q.push({-1,-1});

        int dist=0;
        while(!q.empty())
        {
            i=q.front().first;
            j=q.front().second;
            q.pop();

            if(i==-1){
                dist++;
                if(!q.empty()) q.push({-1,-1});
                continue;
            }

            if(i-1 >=0 && grid[i-1][j]==0 && dist+1<safeD[i-1][j]) {
                safeD[i-1][j]=dist+1;
                q.push({i-1,j});
            }
            if(j-1 >=0 && grid[i][j-1]==0 && dist+1<safeD[i][j-1]) {
                safeD[i][j-1]=dist+1;
                q.push({i,j-1});
            }
            if(i+1 < n && grid[i+1][j]==0 && dist+1<safeD[i+1][j]) {
                safeD[i+1][j]=dist+1;
                q.push({i+1,j});
            }
            if(j+1 < n && grid[i][j+1]==0 && dist+1<safeD[i][j+1]) {
                safeD[i][j+1]=dist+1;
                q.push({i,j+1});
            }
        }
    }
    int maximumSafenessFactor(vector<vector<int>>& grid) 
    {
        int i,j,n=grid.size();
        vector<vector<int>> safeD(n,vector<int>(n,INT_MAX));

        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                if(grid[i][j]==1) safeD[i][j]=0;
            }
        }       

        queue<pair<int,int>> q;

        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                if(grid[i][j]==1) relax(i,j,safeD,grid,q);
            }
        }

        priority_queue<pair<int,pair<int,int>>> pq;
        pq.push({safeD[0][0],{0,0}});
        grid[0][0]=-1;

        int safe;
        while(!pq.empty())
        {
            int val=pq.top().first;
            i=pq.top().second.first;
            j=pq.top().second.second;
            if(i==n-1 && j==n-1) return val;

            pq.pop();
            // right and bottom i+1,j+1

            if(i-1 >= 0 && grid[i-1][j]!=-1){
                safe=safeD[i-1][j];
                grid[i-1][j]=-1;
                pq.push({min(safe,val),{i-1,j}});
            }

            if(j-1 >= 0 && grid[i][j-1]!=-1){
                safe=safeD[i][j-1];
                grid[i][j-1]=-1;
                pq.push({min(safe,val),{i,j-1}});
            }

            if(i+1 < n && grid[i+1][j]!=-1){
                safe=safeD[i+1][j];
                grid[i+1][j]=-1;
                pq.push({min(safe,val),{i+1,j}});
            }

            if(j+1 < n && grid[i][j+1]!=-1){
                safe=safeD[i][j+1];
                grid[i][j+1]=-1;
                pq.push({min(val,safe),{i,j+1}});
            }
        }
        return 0;
    }
};
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