<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 {
private:
    typedef pair<int, pair<int, int>> ppi;

    int n;
    vector<int> delR = {-1, 0, 1, 0}, delC = {0, 1, 0, -1};

    bool isValid(int row, int col){
        return row >= 0 && row < n && col >= 0 && col < n;
    }

    void bfs(vector<vector<int>>& grid,  vector<vector<int>>& dist){
        // Dijsktra Algo - Shortest path from thief to each cell;

        set<ppi> st;
        
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j]){
                    st.insert({0, {i, j}});
                    dist[i][j] = 0;
                } 
            }
        }

        while(!st.empty()){
            auto it = *(st.begin());
            int row = it.second.first;
            int col = it.second.second;
            int dis = it.first;
            st.erase(it);

            for(int i = 0; i < 4; i++){
                int nRow = row + delR[i];
                int nCol = col + delC[i];
                
                int totalD = 1 + dis;
                if(isValid(nRow, nCol) && totalD < dist[nRow][nCol]){
                    int adjD = dist[nRow][nCol];

                    if(adjD != INT_MAX) st.erase({adjD, {nRow, nCol}});

                    dist[nRow][nCol] = totalD;
                    st.insert({totalD, {nRow, nCol}});
                } 
            }
        }
    }

    int maxSafe(vector<vector<int>>& dist){
        // Dijsktra Algo using maxHeap

        priority_queue<ppi> maxH;
        vector<vector<int>> vis(n, vector<int>(n, 0));

        maxH.push({dist[0][0], {0, 0}});
        vis[0][0] = 1;

        while(!maxH.empty()){
            auto it = maxH.top();
            maxH.pop();
            int safeD = it.first;
            int row = it.second.first;
            int col = it.second.second;

            if(row == n-1 && col == n-1) return safeD;

            for(int i = 0; i < 4; i++){
                int nRow = row + delR[i];
                int nCol = col + delC[i];

                if(isValid(nRow, nCol) && !vis[nRow][nCol]){
                    int s = dist[nRow][nCol];
                    int safe = min(s, safeD);
                    vis[nRow][nCol] = 1;
                    maxH.push({safe, {nRow, nCol}});
                }
            }
        }

        return -1;
    }

public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        // Solved By KS

        n = grid.size();

        if(grid[0][0] || grid[n-1][n-1]) return 0;
        vector<vector<int>> dist(n, vector<int>(n, INT_MAX));

        bfs(grid, dist);

        return maxSafe(dist);
    }
};
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