<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 isP(int r, int c, int &safe, vector<vector<int>> &grid, vector<vector<bool>> &isVis){
        if(grid[r][c] < safe) return false;
        
        int n = grid.size();

        isVis[r][c] = true;

        if(r == (n-1) and c == (n-1)) return true;

        vector<int> dr = {-1, 1, 0, 0};
        vector<int> dc = {0, 0, 1, -1};

        for(int i = 0; i < 4; i++){
            int nr = r + dr[i], nc = c + dc[i];
            if(nr < n and nc < n and nr >= 0 and nc >= 0 and !isVis[nr][nc]){
                if(isP(nr, nc, safe, grid, isVis)) return true;
            }
        }

        return false;
    }
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
        int n = grid.size();
        if(n == 1) return 0;
        if(grid[0][0] || grid[n - 1][n - 1]) return 0;
        queue<pair<int, int>> q;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j]){
                    q.push({i, j});
                    grid[i][j] = 0;
                }
                else grid[i][j] = -1;
            }
        }

        // multisource bfs from every cell containing a thief 
        // finding shortest manhattan distance from a given node to one of the cells containing thief

        while(!q.empty()){
            int size = q.size();
            for(int i = 0; i < size; i++){
                auto [r, c] = q.front();
                q.pop();
                if(r + 1 < n and grid[r+1][c] == -1){
                    q.push({r+1, c});
                    grid[r+1][c] = grid[r][c] + 1;
                }
                if(r - 1 >= 0 and grid[r-1][c] == -1){
                    q.push({r-1, c});
                    grid[r-1][c] = grid[r][c] + 1;
                }
                if(c + 1 < n and grid[r][c+1] == -1){
                    q.push({r, c+1});
                    grid[r][c+1] = grid[r][c] + 1;
                }
                if(c - 1 >= 0 and grid[r][c-1] == -1){
                    q.push({r, c-1});
                    grid[r][c-1] = grid[r][c] + 1;
                }
            }
        }


        int low = 0, high = (2*n) - 2;
        int res = 0;
        while(low <= high){
            int mid = (low + high)/2;
            vector<vector<bool>> isvis(n, vector<bool>(n, false));
            if(isP(0, 0, mid, grid, isvis)){
                res = mid;
                low = mid+1;
            }
            else high = mid-1;
        }

        return res;
    }
};
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