<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:
    
    int dx[4] = {-1, 1, 0, 0};
    int dy[4] = {0, 0, -1, 1};
    
    bool isValid(int x, int y, int n) {
        return (0 <= x && x < n && 0 <= y && y < n);
    }
    
    vector<vector<int>> calculateSafeness(vector<vector<int>>& grid) {
        int n = grid.size();
        queue<pair<int,int>> q;
        vector<vector<int>> safeness(n, vector<int>(n, -1));
        
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < n; ++j) {
                if(grid[i][j] == 1) {
                    q.push({i, j});
                    safeness[i][j] = 0;
                }
            }
        }
        
        while(!q.empty()) {
            auto x = q.front();
            q.pop();
            for(int i = 0; i < 4; ++i) {
                int nx = x.first + dx[i];
                int ny = x.second + dy[i];
                if(isValid(nx, ny, n) && safeness[nx][ny] == -1) {
                    q.push({nx, ny});
                    safeness[nx][ny] = safeness[x.first][x.second] + 1;
                }
            }
        }
        return safeness;
    }
    
    bool check(int mid, vector<vector<int>>& safeness) {
        int n = safeness.size();
        vector<vector<int>> visit(n, vector<int>(n, false));
        queue<pair<int,int>> q;
        if(safeness[0][0] >= mid) {
            q.push({0, 0});
            visit[0][0] = true;
        }
        
        while(!q.empty()) {
            auto x = q.front();
            if(x.first == n - 1 && x.second == n - 1) {
                return true;
            }
            q.pop();
            for(int i = 0; i < 4; ++i) {
                int nx = x.first + dx[i];
                int ny = x.second + dy[i];
                if(isValid(nx, ny, n) && !visit[nx][ny] && safeness[nx][ny] >= mid) {
                    q.push({nx, ny});
                    visit[nx][ny] = true;
                }
            }
        }
        return false;
    }
    
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        vector<vector<int>> safeness = calculateSafeness(grid);
        
        int low = 0, high = 400;
        int ans;
        while(low <= high) {
            int mid = (low + high) / 2;
            if(check(mid, safeness)) {
                low = mid + 1;
                ans = mid;
            }
            else {
                high = mid - 1;
            }
        }
        return ans;
    }
};
288058