<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 {
    const vector<int> dx = {-1, 0, 1, 0};
    const vector<int> dy = {0, 1, 0, -1};

    vector<vector<int>> getSafenessFactor(vector<vector<int>> &grid) {

        int n = grid.size(), m = grid[0].size();

        vector<vector<int>> safeNessFactor(n, vector<int>(m, 1e9));
        queue<pair<int,int>> q;

        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if(grid[i][j] == 1) {
                    safeNessFactor[i][j] = 0;
                    q.push({i, j});
                } 
            }
        }


        while(!q.empty()) {
            auto [i, j] = q.front();
            q.pop();

            for(int k=0; k<4; k++) {
                int _i = i + dx[k], _j = j + dy[k];
                if(_i >= 0 && _i < n && _j >= 0 && _j < m && safeNessFactor[_i][_j] == 1e9) {
                    safeNessFactor[_i][_j] = safeNessFactor[i][j] + 1;
                    q.push({_i, _j});
                }
            }
            
        }
        return safeNessFactor;
    }

    int getMaxSafenessFactor(vector<vector<int>> &safenessFactor) {
        int n = safenessFactor.size(), m = safenessFactor[0].size();
        
        deque<pair<int, pair<int,int>>> q;
        vector<vector<bool>> vis(n, vector<bool> (m, false));

        vis[0][0] = true;
        q.push_back({safenessFactor[0][0], {0, 0}});

        int currentSafeness = safenessFactor[0][0];

        while(!q.empty()) {
            int safeNess = q.front().first;
            auto [i, j] = q.front().second;
            q.pop_front();

            currentSafeness = min(currentSafeness, safeNess);

            if(i == n-1 && j == m-1) {
                return currentSafeness;
            }

            for(int k=0; k<4; k++) {
                int _i = i + dx[k], _j = j + dy[k];
                if(_i >= 0 && _i < n && _j >= 0 && _j < m && vis[_i][_j] == false) {
                    vis[_i][_j] = true;
                    if(safenessFactor[_i][_j] < currentSafeness) {
                        q.push_back({min(safenessFactor[_i][_j], currentSafeness), {_i, _j}});
                    } else {
                        q.push_front({currentSafeness, {_i, _j}});
                    }
                }
            }
        }
        return -1;
    }

public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        
        vector<vector<int>> safenessFactor = getSafenessFactor(grid);
        return getMaxSafenessFactor(safenessFactor);
    }
};
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