<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:

    typedef pair<int,int> pii;
    typedef vector<vector<int>> vvi;
    typedef vector<int> vi;

    vi rows = {-1, 0, 1, 0};
    vi cols = {0, 1, 0, -1};

    bool pathPossible(vvi grid, int reqSafeness){

        int n = grid.size();

        if(grid[0][0] < reqSafeness)
            return false;

        queue<pii> q;
        q.push({0,0});

        grid[0][0] = -1;

        while(!q.empty()){

            int i = q.front().first; 
            int j = q.front().second; 
            q.pop();

            if(i == n-1 && j == n-1)
                return true;

            for(int nbr = 0; nbr<4; nbr++){

            int i_ = i + rows[nbr];
            int j_ = j + cols[nbr];

            // out of. bound or already visited cells -->
                if( (i_ >= n) || (i_ < 0) || (j_ >= n)
                    || (j_ < 0) || (grid[i_][j_] == -1) ) 
                    continue;

                if(grid[i_][j_] >= reqSafeness){

                    q.push({i_, j_});
                    grid[i_][j_] = -1;
                    
                }

            }
            
        }
        
        return false;
    }



    int maximumSafenessFactor(vector<vector<int>>& grid) {

        int n = grid.size();

        vvi visited(n, vi(n, false));

        // prepopulate distances from cells to nearest theifs -->
        queue<pii> q;

        // push theifs in queue -->
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){

                if(grid[i][j] == 1){
                    
                    q.push({i,j});
                    visited[i][j] = true;
                    grid[i][j] = 0;
                    
                }
                
            }
        }

        // apply multi source BFS -->
        int lvl = 1;
        while(!q.empty()){

            int queSize = q.size();
            while(queSize--){

                int i = q.front().first; 
            int j = q.front().second; 
            q.pop();

                for(int nbr = 0; nbr<4; nbr++){

                    int i_ = i + rows[nbr];
                    int j_ = j + cols[nbr];

                    // out of. bound or already visited cells -->
                    if( (i_ >= n) || (i_ < 0) || (j_ >= n)
                        || (j_ < 0) || (visited[i_][j_]) ) 
                        continue;

                    // fill manhatten distance and mark cells visited -->
                    grid[i_][j_] = lvl;
                    visited[i_][j_] = true;

                    // push in queue -->
                    q.push({i_, j_});
                    
                }
                
            }
            lvl++;
        }

        // Apply binary search -->
        int left = 0;
        int right = 1e9;

        int maxSafeness = INT_MIN;

        while(left <= right){

            int mid = (left + right)/2;

            if(pathPossible(grid, mid)){
                
                maxSafeness = max(maxSafeness, mid);
                left = mid + 1;
            }
            else{
                right = mid - 1;
            }
            
        }

        return maxSafeness;
            
    }
};
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