<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:
#     def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
#         m = len(grid)
#         if grid[0][0]==1 or grid[-1][-1]==1:
#             return 0
#         directs = [[0,1],[0,-1],[1,0],[-1,0]]
#         theifs = []
#         for i in range(m):
#             for j in range(m):
#                 v = grid[i][j]
#                 if v==1:
#                     theifs.append([i,j])
#                     grid[i][j]='*'
#         seen = {}
#         for i,j in theifs:
#             stack = [[i,j]]
#             # temp  = stack
#             while stack:
#                 temp = stack
#                 stack = []
#                 for x,y in temp:
#                     for drv in directs:
#                         x1,y1 = x+drv[0],y+drv[1]
#                         if 0<=x1<m and 0<=y1<m and grid[x1][y1]!='*' :
#                             dist = abs(x1-i)+abs(y1-j)
#                             if (x1,y1) not in seen or seen[x1,y1]>dist:
#                                 seen[(x1,y1)]=dist
#                                 v = grid[x1][y1]
                                
#                                 # if v==0 or dist<=v:
#                                 stack.append([x1,y1])
#                                 grid[x1][y1]=dist
#         # print(grid)
#         res = -math.inf
#         stack = [[0,0,grid[0][0]]]
#         seen = {(0,0):grid[0][0]}
#         while stack:
#             temp = stack
#             stack = []
#             for i,j,dist in temp:
#                 for drv in directs:
#                     x1,y1 = i+drv[0],j+drv[1]
#                     if 0<=x1<m and 0<=y1<m and grid[x1][y1]!='*':
#                         newDist = min(dist, grid[x1][y1])

#                         if (x1,y1) not in seen or seen[(x1,y1)]<newDist:
#                             seen[(x1,y1)]=newDist
#                             # print(f'x1:{x1} y1:{y1} dist:{newDist} gridxy:{grid[x1][y1]}')
#                             if res>=newDist:
#                                 continue                            
#                             if x1==m-1 and y1==m-1:
#                                 res = max(newDist, res)
#                                 if res==grid[0][0]:
#                                     return res

#                             else:
#                                 stack.append((x1,y1,newDist))

#         return 0 if res == -math.inf else res
            



class Solution:

    # Directions for moving to neighboring cells: right, left, down, up
    dir = [(0, 1), (0, -1), (1, 0), (-1, 0)]

    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        n = len(grid)

        multi_source_queue = deque()
        # Mark thieves as 0 and empty cells as -1, and push thieves to the queue
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    # Push thief coordinates to the queue
                    multi_source_queue.append((i, j))
                    # Mark thief cell with 0
                    grid[i][j] = 0
                else:
                    # Mark empty cell with -1
                    grid[i][j] = -1

        # Calculate safeness factor for each cell using BFS
        while multi_source_queue:
            size = len(multi_source_queue)
            while size > 0:
                curr = multi_source_queue.popleft()
                # Check neighboring cells
                for d in self.dir:
                    di, dj = curr[0] + d[0], curr[1] + d[1]
                    val = grid[curr[0]][curr[1]]
                    # Check if the cell is valid and unvisited
                    if self.isValidCell(grid, di, dj) and grid[di][dj] == -1:
                        # Update safeness factor and push to the queue
                        grid[di][dj] = val + 1
                        multi_source_queue.append((di, dj))
                size -= 1

        # Binary search for maximum safeness factor
        start, end, res = 0, 0, -1
        for i in range(n):
            for j in range(n):
                # Set end as the maximum safeness factor possible
                end = max(end, grid[i][j])

        while start <= end:
            mid = start + (end - start) // 2
            if self.isValidSafeness(grid, mid):
                # Store valid safeness and search for larger ones
                res = mid
                start = mid + 1
            else:
                end = mid - 1

        return res
    
    # Check if a given cell lies within the grid
    def isValidCell(self, grid, i, j) -> bool:
        n = len(grid)
        return 0 <= i < n and 0 <= j < n

    # Check if a path exists with given minimum safeness value
    def isValidSafeness(self, grid, min_safeness) -> bool:
        n = len(grid)

        # Check if the source and destination cells satisfy minimum safeness
        if grid[0][0] < min_safeness or grid[n - 1][n - 1] < min_safeness:
            return False

        traversal_queue = deque([(0, 0)])
        visited = [[False] * n for _ in range(n)]
        visited[0][0] = True

        # Breadth-first search to find a valid path
        while traversal_queue:
            curr = traversal_queue.popleft()
            if curr[0] == n - 1 and curr[1] == n - 1:
                return True  # Valid path found

            # Check neighboring cells
            for d in self.dir:
                di, dj = curr[0] + d[0], curr[1] + d[1]
                # Check if the neighboring cell is valid and unvisited
                if self.isValidCell(grid, di, dj) and not visited[di][dj] and grid[di][dj] >= min_safeness:
                    visited[di][dj] = True
                    traversal_queue.append((di, dj))

        return False  # No valid path found


class Solution:

    # Directions for moving to neighboring cells: right, left, down, up
    dir = [(0, 1), (0, -1), (1, 0), (-1, 0)]

    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        n = len(grid)

        multi_source_queue = deque()
        # Mark thieves as 0 and empty cells as -1, and push thieves to the queue
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    # Push thief coordinates to the queue
                    multi_source_queue.append((i, j))
                    # Mark thief cell with 0
                    grid[i][j] = 0
                else:
                    # Mark empty cell with -1
                    grid[i][j] = -1

        # Calculate safeness factor for each cell using BFS
        while multi_source_queue:
            size = len(multi_source_queue)
            while size > 0:
                curr = multi_source_queue.popleft()
                # Check neighboring cells
                for d in self.dir:
                    di, dj = curr[0] + d[0], curr[1] + d[1]
                    val = grid[curr[0]][curr[1]]
                    # Check if the cell is valid and unvisited
                    if self.isValidCell(grid, di, dj) and grid[di][dj] == -1:
                        # Update safeness factor and push to the queue
                        grid[di][dj] = val + 1
                        multi_source_queue.append((di, dj))
                size -= 1
        
        # Priority queue to prioritize cells with higher safeness factor
        pq = [[-grid[0][0], 0, 0]] # [maximum_safeness_till_now, x-coordinate, y-coordinate]
        grid[0][0] = -1 # Mark the source cell as visited

        # BFS to find the path with maximum safeness factor
        while pq:
            safeness, i, j = heapq.heappop(pq)
            
            # If reached the destination, return safeness factor
            if i == n - 1 and j == n - 1:
                return -safeness
            
            # Check neighboring cells
            for d in self.dir:
                di, dj = i + d[0], j + d[1]
                # Check if the neighboring cell is valid and unvisited
                if self.isValidCell(grid, di, dj) and grid[di][dj] != -1:
                    heapq.heappush(pq, [-min(-safeness, grid[di][dj]), di, dj])
                    grid[di][dj] = -1

        return -1
    
    # Check if a given cell lies within the grid
    def isValidCell(self, grid, i, j) -> bool:
        n = len(grid)
        return 0 <= i < n and 0 <= j < n



class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        m = len(grid)
        if grid[0][0]==1 or grid[m-1][m-1]==1:
            # print('yes')
            return 0
        directs = [[0,1],[0,-1],[1,0],[-1,0]]
        theifs = []
        # sum = 0
        for i in range(m):
            for j in range(m):
                v = grid[i][j]
                if v==1:
                    theifs.append([i,j])
                    grid[i][j]='*'
                # sum+=1
        # print(sum)
        seen = {}
        ref = deque()
        for i,j in theifs:
            ref.append([i,j,0])
        # for i,j in theifs:
        #     stack = [[i,j]]
        #     # temp  = stack
        while ref:
            # temp = stack
            # stack = []
            x,y,d = ref.popleft()
            for drv in directs:
                x1,y1 = x+drv[0],y+drv[1]
                if 0<=x1<m and 0<=y1<m and grid[x1][y1]!='*':
                    dist = d+1
                    if (x1,y1) not in seen:
                        seen[(x1,y1)]=dist
                        v = grid[x1][y1]
                        
                        # if v==0 or dist<=v:
                        ref.append([x1,y1,dist])
                        grid[x1][y1]=dist
            # if ref[i,j]==[]:
            #     del ref[i,j]
        # while len(ref)>0:
        #     # temp = stack
        #     # stack = []
        #     for i,j in theifs:
        #         if (i,j) not in ref:
        #             continue
        #         x,y = ref[i,j].pop()
        #         for drv in directs:
        #             x1,y1 = x+drv[0],y+drv[1]
        #             if 0<=x1<m and 0<=y1<m and grid[x1][y1]!='*':
        #                 dist = abs(x1-i)+abs(y1-j)
        #                 if (x1,y1) not in seen or seen[x1,y1]>dist:
        #                     seen[(x1,y1)]=dist
        #                     v = grid[x1][y1]
                            
        #                     # if v==0 or dist<=v:
        #                     ref[i,j].append([x1,y1])
        #                     grid[x1][y1]=dist
        #         if ref[i,j]==[]:
        #             del ref[i,j]
        # print(grid)
        res = -math.inf
        q = [[-grid[0][0],0,0]]
        dists = {(0,0):grid[0][0]}
        while q:
            # temp = stack
            # stack = []
            dist,i,j= heapq.heappop(q)
            dist = -dist
            for drv in directs:
                x1,y1 = i+drv[0],j+drv[1]
                if 0<=x1<m and 0<=y1<m and grid[x1][y1]!='*':
                    if dist<grid[x1][y1]:
                        newDist = dist
                    else:
                        newDist = grid[x1][y1]
                    # newDist = min(dist, grid[x1][y1])

                    if (x1,y1) not in dists or dists[(x1,y1)]<newDist:
                        if x1==m-1 and y1==m-1:
                            # res = max(newDist, res)
                            return newDist
                        dists[(x1,y1)]=newDist
                        heapq.heappush(q,(-newDist,x1,y1))
                        # q.append((-newDist,x1,y1))
                        # print(f'x1:{x1} y1:{y1} dist:{newDist} gridxy:{grid[x1][y1]}')
                        # if res>=newDist:
                        #     continue                            

                            # if res==grid[0][0]:
                            #     return res
        return 0 if res == -math.inf else res

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