<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>

from collections import deque

class Solution:
    def maximumSafenessFactor(self, grid):
        # Calculate safeness factors for each cell
        safeness = self.calculateSafenessFactors(grid)
        # Find the maximum safeness value from the safeness matrix
        maxSafeness = self.findMaxSafeness(safeness)
        # Perform binary search to find the maximum safeness factor
        return self.findMaxSafenessFactor(grid, safeness, maxSafeness)

    # Method to calculate the safeness factor for each cell using multi-source BFS
    def calculateSafenessFactors(self, grid):
        n = len(grid)
        safeness = [[float('inf')] * n for _ in range(n)]  # Initialize safeness matrix with infinity
        queue = deque()

        # Add all cells with gold to the queue and set their safeness to 0
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    queue.append((i, j))
                    safeness[i][j] = 0

        directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]  # Possible directions for movement

        # Perform BFS to calculate safeness factors
        while queue:
            x, y = queue.popleft()
            for dx, dy in directions:
                newX, newY = x + dx, y + dy
                # Check if the new cell is within bounds and update safeness if necessary
                if 0 <= newX < n and 0 <= newY < n and safeness[newX][newY] > safeness[x][y] + 1:
                    safeness[newX][newY] = safeness[x][y] + 1
                    queue.append((newX, newY))

        return safeness

    # Method to find the maximum safeness value from the safeness matrix
    def findMaxSafeness(self, safeness):
        return max(max(row) for row in safeness)

    # Method to perform binary search to find the maximum safeness factor
    def findMaxSafenessFactor(self, grid, safeness, maxSafeness):
        left, right, result = 0, maxSafeness, 0

        # Binary search to find the maximum safeness factor
        while left <= right:
            mid = left + (right - left) // 2
            # Check if the current mid value can be achieved
            if self.canAchieveSafeness(grid, safeness, mid):
                result = mid  # Update result if the current mid is achievable
                left = mid + 1
            else:
                right = mid - 1

        return result

    # Method to check if a path with the given safeness factor exists using BFS
    def canAchieveSafeness(self, grid, safeness, threshold):
        n = len(grid)
        if safeness[0][0] < threshold:  # If the starting cell's safeness is less than the threshold
            return False

        queue = deque([(0, 0)])  # Start BFS from the top-left corner
        visited = [[False] * n for _ in range(n)]
        visited[0][0] = True

        directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]  # Possible directions for movement

        # Perform BFS to check if the bottom-right corner can be reached
        while queue:
            x, y = queue.popleft()
            if x == n - 1 and y == n - 1:  # If we reach the bottom-right corner
                return True

            for dx, dy in directions:
                newX, newY = x + dx, y + dy
                # Check if the new cell is within bounds and its safeness is above the threshold
                if 0 <= newX < n and 0 <= newY < n and not visited[newX][newY] and safeness[newX][newY] >= threshold:
                    queue.append((newX, newY))
                    visited[newX][newY] = True

        return False
 
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