Find the Safest Path in a Grid
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2800-2899/2812.Find%20the%20Safest%20Path%20in%20a%20Grid/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2800-2899/2812.Find%20the%20Safest%20Path%20in%20a%20Grid/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2800-2899/2812.Find%20the%20Safest%20Path%20in%20a%20Grid/images/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>
Breadth-First Search; Union Find; Array; Binary Search; Matrix; Heap (Priority Queue)
type unionFind struct {
	p []int
	n int
}

func newUnionFind(n int) *unionFind {
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	return &unionFind{p, n}
}

func (uf *unionFind) find(x int) int {
	if uf.p[x] != x {
		uf.p[x] = uf.find(uf.p[x])
	}
	return uf.p[x]
}

func (uf *unionFind) union(a, b int) bool {
	if uf.find(a) == uf.find(b) {
		return false
	}
	uf.p[uf.find(a)] = uf.find(b)
	uf.n--
	return true
}

func maximumSafenessFactor(grid [][]int) int {
	n := len(grid)
	if grid[0][0] == 1 || grid[n-1][n-1] == 1 {
		return 0
	}
	q := [][2]int{}
	dist := make([][]int, n)
	const inf = 1 << 30
	for i := range dist {
		dist[i] = make([]int, n)
		for j := range dist[i] {
			dist[i][j] = inf
		}
	}
	for i := 0; i < n; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] == 1 {
				dist[i][j] = 0
				q = append(q, [2]int{i, j})
			}
		}
	}
	dirs := [5]int{-1, 0, 1, 0, -1}
	for len(q) > 0 {
		p := q[0]
		q = q[1:]
		i, j := p[0], p[1]
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < n && y >= 0 && y < n && dist[x][y] == inf {
				dist[x][y] = dist[i][j] + 1
				q = append(q, [2]int{x, y})
			}
		}
	}
	t := [][3]int{}
	for i := 0; i < n; i++ {
		for j := 0; j < n; j++ {
			t = append(t, [3]int{dist[i][j], i, j})
		}
	}
	sort.Slice(t, func(i, j int) bool {
		return t[i][0] > t[j][0]
	})
	uf := newUnionFind(n * n)
	for _, p := range t {
		d, i, j := p[0], p[1], p[2]
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < n && y >= 0 && y < n && dist[x][y] >= d {
				uf.union(i*n+j, x*n+y)
			}
		}
		if uf.find(0) == uf.find(n*n-1) {
			return d
		}
	}
	return 0
}