<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>

# 1631. Path With Minimum Effort
# n x n grid
#Time: n * n * log(n * n)  Each cell is potentially inserted into the priority queue, and the queue operations take O(log(n)) time.
#Space: n * n 
class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        if grid[0][0] == 1 or grid[-1][-1] == 1: return 0
        
        #先计算Manhattan distance，下面就可以focus on **minimum manhattan distance from any cell
        #542. 01 Matrix
        n = len(grid)  # n * n
        
        queue = deque()
        for i in range(n):
            for j in range(n):
                if(grid[i][j] == 1):
                    grid[i][j] = 0
                    queue.append([i, j])
    
                elif(grid[i][j] == 0):
                    grid[i][j] = math.inf
                    
        distance = 0            
        delta = [[1, 0], [-1, 0], [0, 1], [0, -1]]            
        while(queue):
            
            distance += 1
            for _ in range(len(queue)):
                x, y = queue.popleft()

                for dx, dy in delta:
                    newX = x + dx
                    newY = y + dy

                    if(newX < 0 or newX > n - 1 or newY < 0 or newY > n - 1): continue
                    if(grid[newX][newY] != math.inf): continue

                    grid[newX][newY] = distance
                    queue.append([newX, newY])
                    
        #现在grid里面已经都是safeness factor/manhattan distance了
                
        #Dijkstra
        safeness = [[0] * n for _ in range(n)] #Return the maximum safeness factor -> initial as 0 #和heap相反
        safeness[0][0] = grid[0][0]
        
        maxHeap = [(-grid[0][0], 0, 0)] #Return the maximum safeness factor 
        
        while(maxHeap):
            curSafe, x, y = heapq.heappop(maxHeap)
            curSafe = -curSafe
            
            if((x, y) == (n - 1, n - 1)): return curSafe
            
            # The safeness factor is defined as the minimum manhattan distance 
            if(curSafe < safeness[x][y]): continue
                
            for dx, dy in delta:
                newX = x + dx
                newY = y + dy
                
                if(newX < 0 or newX > n - 1 or newY < 0 or newY > n - 1): continue
                    
                #newX newY找到了，现在到newSafe了
                ##直接和grid[newX][newY]对比，因为grid里面已经是计算好了的Manhattan distance
                #现在是找minimum manhattan distance
                newSafe = min(curSafe, grid[newX][newY]) 
                
                if(newSafe > safeness[newX][newY]): #Return the maximum safeness factor 
                    safeness[newX][newY] = newSafe
                    heapq.heappush(maxHeap, (-newSafe, newX, newY))

        return safeness[n-1][n-1]

23581