<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:
    vector<int>rowDir = {-1, 1, 0, 0};
    vector<int>colDir = {0, 0, -1, 1};
    bool isValid(vector<vector<bool>>&visited, int i, int j)
    {
        if (i < 0 || j < 0 || i == visited.size() || j == visited[0].size() || visited[i][j])
            return false;
        return true;
    }
    //============================================================================================================
    bool isSafe(vector<vector<int>>&distToTheif, int safeDist) //check the validity of safenessFactor
    {
        int n = distToTheif.size();
        queue<pair<int, int>>q;
        if (distToTheif[0][0] < safeDist) return false;
        q.push({0, 0});
        vector<vector<bool>>visited(n, vector<bool>(n, false));
        visited[0][0] = true;
        while(!q.empty())
        {
            int currRow = q.front().first, currCol = q.front().second;
            q.pop();
            if (currRow == n - 1 && currCol == n - 1) return true;
            for (int dirIdx = 0; dirIdx < 4; dirIdx++)
            {
                int newRow = currRow + rowDir[dirIdx];
                int newCol = currCol + colDir[dirIdx];
                if (isValid(visited, newRow, newCol))
                {
                    if (distToTheif[newRow][newCol] < safeDist) continue;
                    visited[newRow][newCol] = true;
                    q.push({newRow, newCol});
                }
            }
        }
        return false;
    }
    //===========================================================================================================
    int maximumSafenessFactor(vector<vector<int>>& grid) 
    {
        int n = grid.size();
        queue<pair<int, int>>q;
        vector<vector<bool>>visited(n, vector<bool>(n, false));
        vector<vector<int>>distToTheif(n, vector<int>(n, -1));
        //========================================================================
        //Add all the theives in current queue
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (grid[i][j] == 1) 
                {
                    visited[i][j] = true;
                    q.push({i, j});
                }
            }
        }
        //=============================================================================
        //BFS to fill the "DistToTheif" 2D array
        int dist = 0;
        while(!q.empty())
        {
            int size = q.size();
            while(size--)
            {
                int currRow = q.front().first, currCol = q.front().second;
                q.pop();
                distToTheif[currRow][currCol] = dist;
                for (int dirIdx = 0; dirIdx < 4; dirIdx++)
                {
                    int newRow = currRow + rowDir[dirIdx], newCol = currCol + colDir[dirIdx];
                    if (!isValid(visited, newRow, newCol)) continue;
                    
                    visited[newRow][newCol] = true;
                    q.push({newRow, newCol});
                }
            }
            dist++;
        }
        //==================================================================================
        //BINARY SEARCH
        int low = 0, high = INT_MAX;
        int ans = 0;
        while(low <= high)
        {
            int mid = low + (high - low) / 2;
            if (isSafe(distToTheif, mid))
            {
                ans = mid;
                low = mid + 1;
            }
            else high = mid - 1;
        }
        //=========================================================================================
        return ans;
        
    }
};
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