<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 delrow[4] = {-1,0,+1,0};
    int delcol[4] = {0,+1,0,-1};

    bool isValid(int row,int col,vector<vector<int>> &grid){
        int n = grid.size();
        if(row >= 0 && row < n && col >= 0 &&  col < n){
            return true;
        }
        return false;
    }

    bool helper(int row,int col,int factor,vector<vector<int>> &maxsafe,vector<vector<int>> &grid,vector<vector<int>> &visited){
        int n = grid.size();
        if(maxsafe[row][col] < factor){
            return false;
        }
        if(row == n -1 && col == n -1){
            return true;
        }
        visited[row][col] = 1;

        for(int i = 0;i<4;i++){
            int nrow = row + delrow[i];
            int ncol = col + delcol[i];
            if(isValid(nrow,ncol,grid) && maxsafe[nrow][ncol] >= factor && visited[nrow][ncol] == 0){
                bool b = helper(nrow,ncol,factor,maxsafe,grid,visited);
                if(b){
                    return true;
                }
            }
        }
        return false;
    }

    void calculateMaxSafeness(vector<vector<int>> &maxsafe,vector<vector<int>>& grid){
        int n = maxsafe.size();
        queue<pair<int,int>> q;
        for(int i = 0;i<n;i++){
            for(int j = 0;j<n;j++){
                if(grid[i][j] == 1){
                    maxsafe[i][j] = 0;
                    q.push({i,j});
                }
                else{
                    maxsafe[i][j] = -1;
                }
            }
        }
        while(!q.empty()){
            auto front = q.front();
            int row = front.first;
            int col = front.second;
            q.pop();
            for(int i = 0;i<4;i++){
                int nrow = row + delrow[i];
                int ncol = col + delcol[i];
                if(isValid(nrow,ncol,grid) && maxsafe[nrow][ncol] == -1){
                    maxsafe[nrow][ncol] = maxsafe[row][col] + 1;
                    q.push({nrow,ncol});
                }
            }

        }
    }

    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        int low = 0;
        int high = n - 1;
        vector<vector<int>> maxsafe(n,vector<int>(n));
        calculateMaxSafeness(maxsafe,grid);

        while(low <= high){
            int mid = (high + low)/2;
            vector<vector<int>> visited(n,vector<int>(n,0));
            if(helper(0,0,mid,maxsafe,grid,visited)){
                low = mid + 1;
            }
            else{
                high = mid - 1;
            }
        }
        return high;
    }
};
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