<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 {
    int rows;
    int cols;
    vector<pair<int, int>> directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    bool isValidCell(int x, int y) {
        return (x >= 0 && x < rows && y >= 0 && y < cols);
    }

    bool bfsPathExists(vector<vector<int>>& distFromThief,
                       int& minPermittedSafeFactor) {
        if (distFromThief[0][0] < minPermittedSafeFactor ||
            distFromThief[rows - 1][cols - 1] < minPermittedSafeFactor) {
            return false;
        }
        vector<vector<bool>> visited(rows, vector<bool>(cols, false));
        queue<pair<int, int>> q;
        q.push({0, 0});
        while (!q.empty()) {
            pair<int, int> p = q.front();
            int x = p.first;
            int y = p.second;
            q.pop();
            visited[x][y] = true;
            if (x == rows - 1 && y == cols - 1) {
                return true;
            }
            for (auto direction : directions) {
                int nextX = x + direction.first;
                int nextY = y + direction.second;
                if (!isValidCell(nextX, nextY) || visited[nextX][nextY] ||
                    distFromThief[nextX][nextY] < minPermittedSafeFactor) {
                    continue;
                }
                q.push(make_pair(nextX, nextY));
            }
        }
        return false;
    }

public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int tr = grid.size();
        int tc = grid[0].size();
        rows = tr;
        cols = tc;
        vector<vector<int>> distFromThief(tr, vector<int>(tc, INT_MAX));
        queue<pair<int, int>> q;
        for (int i = 0; i < tr; i++) {
            for (int j = 0; j < tc; j++) {
                if (grid[i][j] == 1) {
                    q.push(make_pair(i, j));
                }
            }
        }
        int maxSafeFactor = 0;
        int dist = 0;
        while (!q.empty()) {
            int qs = q.size();
            maxSafeFactor = dist;
            while (qs--) {
                pair<int, int> p = q.front();
                int x = p.first;
                int y = p.second;
                q.pop();
                if (isValidCell(x, y) == false || dist >= distFromThief[x][y]) {
                    continue;
                }
                distFromThief[x][y] = dist;
                q.push(make_pair(x + 1, y));
                q.push(make_pair(x - 1, y));
                q.push(make_pair(x, y + 1));
                q.push(make_pair(x, y - 1));
            }
            dist++;
        }
        // cout << "bfs complete" << endl;

        priority_queue<vector<int>>pq;
        pq.push({distFromThief[0][0],0,0});
        while(!pq.empty()) {
            vector<int> curr = pq.top();
            pq.pop();
            if(curr[1]==rows-1 && curr[2]==cols-1) {
                return curr[0];
            }
            for(pair<int, int> p: directions) {
                int x = curr[1] + p.first;
                int y = curr[2] + p.second;
                if(isValidCell(x,y) && distFromThief[x][y]!=-1) {
                    pq.push({min(curr[0], distFromThief[x][y]), x, y});
                    distFromThief[x][y] = -1;
                }
            }
        }
        return -1;
        // int lo = 0;
        // int hi = maxSafeFactor;
        // int mid = lo + (hi - lo) / 2;
        // int ans = 0;
        // while (lo <= hi) {
        //     mid = lo + (hi - lo) / 2;
        //     if (bfsPathExists(distFromThief, mid)) {
        //         ans = mid;
        //         lo = mid + 1;
        //     } else {
        //         hi = mid - 1;
        //     }
        // }
        // return ans;
    }
};
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