<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:
        safeness=[[-1]*len(grid) for _ in range(len(grid[0]))]
        q=[]
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j]==1:
                    #FOR MULTI-SOURCE Q NEED TO MARK VISITED FOR ALL INITIAL SOURCES
                    safeness[i][j]=0
                    q.append((i, j, 0))
        
        while q:
            size = len(q)
            while size>0:
                cur=q.pop(0)
                safety=cur[2]
                safeness[cur[0]][cur[1]]=safety
                for dir in [(1,0), (-1,0),(0,-1),(0,1)]:
                    di=cur[0]+dir[0]
                    dj=cur[1]+dir[1]
                    if 0<=di<=len(grid)-1 and 0<=dj<=len(grid[0])-1 and safeness[di][dj]==-1:
                        #FOR MULTI-SOURCE QUEUE BFS NEED TO MARK VISITED/NEW DIST BEFORE QUEUE POP
                        safeness[di][dj]=safety+1
                        q.append((di, dj, safety+1))
                size-=1

        visited=[[-1]*len(grid) for _ in range(len(grid[0]))]
        heap=[]
        heappush(heap, (-safeness[0][0], (0,0)))
        visited[0][0]=safeness[0][0]
        globalmin=safeness[0][0]
        while heap:
            curr=heappop(heap)
            #visited[curr[1][0]][curr[1][1]]=-curr[0]
            for dir in [(1,0),(0,1),(-1,0),(0,-1)]:
                x=curr[1][0]+dir[0]
                y=curr[1][1]+dir[1]

                #THIS IS DIJKSTRA NOT PRIM BECAUSE THE SAFENESS FACTOR
                #OF A PATH IS SIMILAR TO PATH LENGTH IN THAT IT'S
                #SOME PROPERTY OF THE PATH AS A WHOLE
                #only visit a grid if the safeness of its path can be improved.
                #only need to check if visited since it isn't like shortest path
                #the solution can only get worse in the future because it is not a cumulative
                #with possible shorter edges, the "value" of the path is always the least 
                #safe grid in the path
                if 0<=x<len(grid) and 0<=y<len(grid[0]) and visited[x][y]==-1:
                    if x==len(grid)-1 and y==len(grid[0])-1:
                        return min(safeness[x][y], visited[curr[1][0]][curr[1][1]])
                    visited[x][y]=min(safeness[x][y], visited[curr[1][0]][curr[1][1]])
                    heappush(heap, (-visited[x][y], (x,y)))
        return visited[-1][-1]

    #THIS IS NOT BFS, I LABELLED IT INCORRECTLY, THIS IS DFS
    # def BFS(self, safeness, x, y, curr_safe):
    #     safeness[x][y]=curr_safe
    #     for dir in [(1,0), (-1,0),(0,-1),(0,1)]:
    #         new_x=dir[0]+x
    #         new_y=dir[1]+y
    #         if 0<=new_x<len(safeness) and 0<=new_y<len(safeness[0]):
    #             if safeness[new_x][new_y]>curr_safe+1:
    #                 self.BFS(safeness, new_x, new_y, curr_safe+1)
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