<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:
        safe = []
        n = len(grid)
        if grid[0][0] == 1 or grid[n-1][n-1] == 1:
            return 0

        for i in range(n):
            safe.append([])
            for j in range(n):
                safe[i].append(math.inf)
        
        q = []
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    # self.bfs(i, j, safe)
                    q.append(((i,j), (i,j)))
        self.bfs(q, safe, grid)

        return self.dijkstra(safe)

    def dijkstra(self, safe):
        n = len(safe)
        q = [(-1*safe[0][0], (0,0))]
        table = []
        for i in range(n):
            table.append([])
            for j in range(n):
                table[i].append(-math.inf)
        table[0][0] = safe[0][0]

        visited = set()
        processed = set()
        while q:
            c = heapq.heappop(q)
            safeness, i, j = -1*c[0], c[1][0],c[1][1]
            if (i,j) not in visited:
                visited.add((i,j))
                if i == n-1 and j == n-1:
                    break

                options = [(i+1,j), (i-1,j), (i,j+1), (i,j-1)]
                for o in options:
                    if o[0] < 0 or o[1] < 0 or o[0] == n or o[1] == n:
                        continue
                    if o not in visited:
                        next_safeness = min(safeness, safe[o[0]][o[1]])
                        table[o[0]][o[1]] = max(next_safeness, table[o[0]][o[1]])
                        item = (-1*next_safeness, o)
                        if item not in processed:
                            processed.add(item)
                            heapq.heappush(q, item)

        return table[n-1][n-1]


        n = len(safe)
        best = min(safe[0][0], safe[n-1][n-1])
        q = [(0,0)]
        visited, processed = set(), {(0,0)}
        index, end = -1, len(q)-1
        layer_best = best
        final = False
        while q:
            index += 1
            c = q.pop(0)
            if c not in visited:
                visited.add(c)
                ci, cj = c[0], c[1]
                if ci == n-1 and cj == n-1:
                    final = True
                    print("final")
                    break

                layer_best = max(layer_best, safe[ci][cj])
                options = [(ci+1,cj),(ci-1,cj),(ci,cj+1),(ci,cj-1)]
                for o in options:
                    if o[0] < 0 or o[1] < 0 or o[0] == n or o[1] == n:
                        continue
                    if safe[o[0]][o[1]] == 0:
                        processed.add(o)
                        continue
                    if o not in processed and o not in visited:
                        q.append(o)
                        processed.add(o)

            if index == end:
                print("layer_best, best:", layer_best, best)
                best = min(best, layer_best)
                index, end = -1, len(q)-1
                layer_best = 0

        # print("final:", final)
        if not final:
            return 0

        return best

    # being used as a multi source bfs
    def bfs(self, q, safe, grid):
        visited, processed = set(), set()
        while q:

            c = q.pop(0)
            cell, thief = c[0], c[1]
            processed.add(cell)
            if cell not in visited:
                visited.add(cell)
                i, j = cell[0], cell[1]
                ri, rj = thief[0], thief[1]
                val = abs(ri-i) + abs(rj-j)
                safe[i][j] = min(safe[i][j], val)
                options = [(i+1,j), (i-1,j), (i,j+1), (i,j-1)]
                for o in options:
                    if o[0] < 0 or o[1] < 0 or o[0] == len(safe) or o[1] == len(safe):
                        continue
                    root = thief
                    if grid[o[0]][o[1]] == 1:
                        root = o
                    item = (o,root)
                    if o not in processed and o not in visited:
                        q.append(item)
                        processed.add(item)


# if [i][j] == 1 we have a thief (not safe)
# if [i][j] == 0 we have a safe cell

# start is [0][0]
# target is [n-1][n-1]

# safeness factor is the min manhattan distance from any cell to any thief in the grid

# the goal is the maximum safeness factor

# we basically want the value of the safest path

# we must know what's the safeness factor of a cell
# safeness[i][j] = the manhattan distance to the closest thief cell

# we run BFS from a particular root cell
# we keep track of visited
# and we keep track of the original root
# and we add the cells to the queue when they are not visited or processed yet
# and every time we compare the existing value of that cell safeness[i][j] and put the minimum value there

# we iterate through all the cells in the original grid and if they are thiefs we run the bfs algorithm to calculate the safeness factor

# after all that we will be left with a safeness grid

# run dijkstra to optimize the max cost (reverse heap)?

# maybe just run BFS and choose the one with max value
# what if we have edges with same val?

# [2,2,1,0]
# [1,2,1,0]
# [0,1,0,0]
# [0,1,1,0]
# [0,1,1,0]

# val = min(val, max(layer))

# 3,2,1,0
# 1,1,0,0
# 0,1,1,1

#[1, 0, 0]
#[1, 0, 0]
#[0, 1, 1]]

#[0,1,1]
#[0,1,0]
#[0,1,0]

#[1, 0, 0],
#[1, 0, 1],
#[1, 0, 1]]


#[0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0],
#[0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0],
#[0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0],
#[0,0,0,0,0,0,1,0,1,0,0,0,0,1,1,0,0,0,0,0,0,0,0],
#[0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0],
#[0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,1,0,1],
#[0,0,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,0],
#[0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0],
#[0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0],
#[0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,0,0,0],
#[0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1],
#[1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0],
#[0,0,0,0,0,0,0,0,0,1,0,1,0,0,1,0,0,0,0,0,0,0,1],
#[0,0,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0],
#[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
#[0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,1,1,0,0,0],
#[0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0],
#[0,0,1,0,0,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0],
#[0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,1,0,0,0],
#[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
#[0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0],
#[0,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0],
#[0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0]]
7054