<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:
    bool dfs(int i,int j,int leastSafety,vector<vector<int>>& safetyFactor,int n,vector<vector<int>>& vis)
    {
        vis[i][j] = 1;
        if(i== n-1 && j == n-1 && safetyFactor[i][j] >= leastSafety)
        {
            return true;
        }
        if(safetyFactor[i][j] < leastSafety)
        {
            return false;
        }
        int dir_r[] = {1,0,-1,0};
        int dir_c[] = {0,1,0,-1};

        for(int x = 0; x < 4; x++)
            {
                int n_row = i + dir_r[x];
                int n_col = j + dir_c[x];

                if(n_row >= 0 && n_row < n && n_col >= 0 && n_col < n
                  && !vis[n_row][n_col]  && dfs(n_row,n_col,leastSafety,safetyFactor,n,vis) == true)
                 {
                    return true;
                 }
            }

            return false;
    }
    bool possible(int leastSafety,vector<vector<int>>& safetyFactor,int n)
    {
        vector<vector<int>> vis(n+1,vector<int>(n+1,0));
        return dfs(0,0,leastSafety,safetyFactor,n,vis);
    }
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>> safetyFactor(n,vector<int>(n,-1));

        // BFS to calculate the safetyFactor for each cell as minimum manahatten distance 
        //from a cell to any thief in the grid is the safety factor

        queue<pair<pair<int,int>,int>> q;

        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(grid[i][j] == 1)
                {
                    q.push({{i,j},0});
                    safetyFactor[i][j] = 0;
                }
            }
        }
        // O(n*n)

        // BFS
        int dir_r[] = {1,0,-1,0};
        int dir_c[] = {0,1,0,-1};
        while(!q.empty())
        {
            auto node = q.front();
            q.pop();

            for(int i = 0; i < 4; i++)
            {
                int n_row = node.first.first + dir_r[i];
                int n_col = node.first.second + dir_c[i];

                if(n_row >= 0 && n_row < n && n_col >= 0 && n_col < n
                 && safetyFactor[n_row][n_col] == -1)
                 {
                    safetyFactor[n_row][n_col] = node.second + 1;
                    q.push({{n_row,n_col},safetyFactor[n_row][n_col]});
                 }
            }
        }
        // O(n*n) traversed all cell 
        // Space - O(n*n)

        // now we will perform binary search on answer to get
        // maximum safetyFactor to reach n-1,n-1 from 0,0
        int low = 0;
        int high = 2*n; // max possible safety factor as maximum manhatten distance could be 2*(n-1) //
        int ans = -1;
        while(low <= high) // O(log(2*n))
        {
            int mid = low + high >> 1;

            if(possible(mid,safetyFactor,n)) // O(n*n)
            {
                ans = mid;
                low = mid+1;
            }
            else
            {
                high = mid-1;
            }
        }

    // TC - O(n*n) --> BFS Traversal + O(n*n *(log(2*n))) --> dfs for each iteration of binary search
    // ==> O(n*2) + O(n*2*logn)
    // SC - O(n*n)
        return ans;
    }
};
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