<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>

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pair<int,int>> vp;
typedef pair<int, int> cell;
class Solution {
public:
    // conditions:
    // 1. move in adjacent directions
    // 2. safeness factor is minimum manhattan dist to a thief
    //    mahattan distance: |a - x| + |b - y|
    int m_dist(int a, int b, int x, int y) {
        return abs(a - x) + abs(b - y);
    }
    
    vp neighbors(int i, int j, int n) {
        // prioritize down and to the right
        vp neighbors;
        if (i < n-1)
            neighbors.push_back({i+1, j});
        if (j < n-1)
            neighbors.push_back({i, j+1});
        if (i > 0)
            neighbors.push_back({i-1, j});
        if (j > 0)
            neighbors.push_back({i, j-1});
            
        return neighbors;
    }
    
    
    void bfs(vvi& grid, vvi& safeness) {
        int n = grid.size();
        queue<cell> q;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j]) {
                    q.push({i, j});
                    safeness[i][j] = 0;
                }
            }
        }
        // bfs
        while (!q.empty()) {
            auto [x, y] = q.front();
            q.pop();
            for (auto& [nx, ny]: neighbors(x, y, n)) {
                int newDist = safeness[x][y] + 1;
                if (newDist < safeness[nx][ny]) {
                    safeness[nx][ny] = newDist;
                    q.push({nx, ny});
                }
            }
        }
        
    }
    
    int widestPath(vvi& safeness) {
        int n = safeness.size();
        auto cmp = [&safeness](cell const& p1, cell const& p2) {
            return safeness[p1.first][p1.second] < safeness[p2.first][p2.second];
        };
        // max heap for nodes based on their safeness
        priority_queue<cell, vector<cell>, decltype(cmp)> pq(cmp);
        
        // start with 0,0
        vvi visited(n, vi(n, 0));
        // max safeness can only go down from 0,0
        int maxSafe = safeness[0][0];
        
        pq.push({0,0});
        while (!pq.empty()) {
            cell cell = pq.top();
            pq.pop();
            int x = cell.first;
            int y = cell.second;
            // already visited - skip
            if (visited[x][y]) continue;
            visited[x][y] = 1;
            // see if this reduces our max safety
            maxSafe = min(maxSafe, safeness[x][y]);
            if (x == n-1 && y == n-1) {
                // reached target
                return maxSafe;
            }
            for (auto& neighbor : neighbors(cell.first, cell.second, n)) {
                pq.push({neighbor.first, neighbor.second});
            }
        }
        
        return maxSafe;
    }
    
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        // calculate max safe path between (0,0) and (n-1, n-1)
        // calculate safeness factors for every cell
        vvi safeness(n, vi(n, INT_MAX - 1000));
        queue<int> q;
        // bfs from each thief, updating manhattan distances
        bfs(grid, safeness);
        
        return widestPath(safeness);
    }
};
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