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+# 2812. Find the Safest Path in a Grid
+
+[LeetCode Link](https://leetcode.com/problems/find-the-safest-path-in-a-grid/)
+
+Difficulty: Medium
+Topics: Array, Binary Search, Breadth-First Search, Union-Find, Heap (Priority Queue), Matrix
+Acceptance Rate: 50.6%
+
+## Hints
+
+### Hint 1
+
+This is really two problems stitched together. First, notice that the "safeness factor" of a path depends only on how close each cell gets to the *nearest* thief. So before you worry about paths at all, ask: for every cell in the grid, how far is it from the closest thief? Think about which traversal computes shortest distances from *multiple* starting points at once.
+
+### Hint 2
+
+Once you have, for each cell, its Manhattan distance to the nearest thief, you can precompute it with a **multi-source BFS**: seed a queue with *all* thief cells at distance 0 and let the wave expand outward. In a 4-connected grid, the number of BFS steps to reach a cell equals its Manhattan distance to the nearest source — so a single BFS fills in the whole distance grid in O(n²).
+
+### Hint 3
+
+Now the real question: among all paths from `(0,0)` to `(n-1,n-1)`, you want the one whose *minimum* cell-distance is as *large* as possible. This is a classic **maximin / widest-path** problem. Two clean ways to solve it: (a) a Dijkstra-style walk with a **max-heap**, always expanding the cell reachable with the largest bottleneck value; or (b) **binary search** on the answer `k` combined with a BFS/DSU that only walks through cells with distance `≥ k`. Both give the same result.
+
+## Approach
+
+**Step 1 — Distance-to-nearest-thief grid (multi-source BFS).**
+Create a `dist` grid initialized to `-1`. Push every thief cell into a queue with `dist = 0`. Run BFS: when you pop a cell, relax its four neighbors, setting their distance to `current + 1` the first time they're visited. Because every source starts at distance 0 simultaneously, the value that lands on each cell is exactly the Manhattan distance to the *closest* thief. This costs O(n²) time.
+
+**Step 2 — Widest path via a max-heap (Dijkstra variant).**
+We want to maximize the minimum `dist` value encountered along a path. Define `safe[r][c]` = the best achievable path-safeness when arriving at `(r, c)`. Start at `(0,0)` with `safe[0][0] = dist[0][0]` and push it onto a max-heap keyed by safeness.
+
+Repeatedly pop the cell with the **largest** current safeness (this greedy choice is what makes Dijkstra correct here). For each neighbor, the safeness of extending the path is `min(current safeness, dist[neighbor])`. If that value improves the neighbor's recorded `safe`, update it and push it. The first time we pop `(n-1, n-1)`, its safeness is the answer — greedy pop order guarantees we've found its optimal bottleneck.
+
+**Why it works.** A path's safeness is a *bottleneck* (minimum along the path), and the max-heap always finalizes the cell with the current best bottleneck first, exactly like Dijkstra finalizes the shortest-distance node first. Substituting `min` for `+` and "max-heap" for "min-heap" turns shortest-path into widest-path.
+
+**Worked example.** For `grid = [[0,0,1],[0,0,0],[0,0,0]]` the only thief is at `(0,2)`. The `dist` grid becomes:
+
+```
+2 1 0
+3 2 1
+4 3 2
+```
+
+Starting at `(0,0)` with safeness 2, we can route down and along cells whose distance never drops below 2, reaching `(2,2)` with safeness 2 — the answer.
+
+## Complexity Analysis
+
+Time Complexity: O(n² log n) — the BFS is O(n²); the heap walk visits each of the n² cells and performs O(log n²) = O(log n) heap operations per push.
+Space Complexity: O(n²) — the `dist` and `safe` grids plus the heap, all bounded by the number of cells.
+
+## Edge Cases
+
+- **Start or destination is a thief** (`grid[0][0] == 1` or `grid[n-1][n-1] == 1`): `dist` at that cell is 0, so every path through it has safeness 0 — the answer is 0. Example 1 is exactly this case.
+- **Every cell is a thief**: all distances are 0, answer is 0.
+- **`n == 1`** (single cell): the start *is* the destination; the answer is `dist[0][0]`, which is 0 since the lone cell must contain the sole thief (the constraint guarantees at least one thief).
+- **Sparse thieves in a large grid**: distances grow large; make sure the heap comparison maximizes safeness and that `safe` values are only overwritten when strictly improved, to avoid redundant work and infinite loops.
+- **Don't forget** cells containing thieves are still walkable — you may pass through a thief cell, it just forces the path safeness to 0.
diff --git a/problems/2812-find-the-safest-path-in-a-grid/problem.md b/problems/2812-find-the-safest-path-in-a-grid/problem.md
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+---
+number: "2812"
+frontend_id: "2812"
+title: "Find the Safest Path in a Grid"
+slug: "find-the-safest-path-in-a-grid"
+difficulty: "Medium"
+topics:
+  - "Array"
+  - "Binary Search"
+  - "Breadth-First Search"
+  - "Union-Find"
+  - "Heap (Priority Queue)"
+  - "Matrix"
+acceptance_rate: 5061.4
+is_premium: false
+created_at: "2026-07-01T05:03:26.509843+00:00"
+fetched_at: "2026-07-01T05:03:26.509843+00:00"
+link: "https://leetcode.com/problems/find-the-safest-path-in-a-grid/"
+date: "2026-07-01"
+---
+
+# 2812. Find the Safest Path in a Grid
+
+You are given a **0-indexed** 2D matrix `grid` of size `n x n`, where `(r, c)` represents:
+
+  * A cell containing a thief if `grid[r][c] = 1`
+  * An empty cell if `grid[r][c] = 0`
+
+
+
+You are initially positioned at cell `(0, 0)`. In one move, you can move to any adjacent cell in the grid, including cells containing thieves.
+
+The **safeness factor** of a path on the grid is defined as the **minimum** manhattan distance from any cell in the path to any thief in the grid.
+
+Return _the**maximum safeness factor** of all paths leading to cell _`(n - 1, n - 1)`_._
+
+An **adjacent** cell of cell `(r, c)`, is one of the cells `(r, c + 1)`, `(r, c - 1)`, `(r + 1, c)` and `(r - 1, c)` if it exists.
+
+The **Manhattan distance** between two cells `(a, b)` and `(x, y)` is equal to `|a - x| + |b - y|`, where `|val|` denotes the absolute value of val.
+
+ 
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2023/07/02/example1.png)
+    
+    
+    **Input:** grid = [[1,0,0],[0,0,0],[0,0,1]]
+    **Output:** 0
+    **Explanation:** All paths from (0, 0) to (n - 1, n - 1) go through the thieves in cells (0, 0) and (n - 1, n - 1).
+    
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2023/07/02/example2.png)
+    
+    
+    **Input:** grid = [[0,0,1],[0,0,0],[0,0,0]]
+    **Output:** 2
+    **Explanation:** 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.
+    
+
+**Example 3:**
+
+![](https://assets.leetcode.com/uploads/2023/07/02/example3.png)
+    
+    
+    **Input:** grid = [[0,0,0,1],[0,0,0,0],[0,0,0,0],[1,0,0,0]]
+    **Output:** 2
+    **Explanation:** 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.
+    
+
+ 
+
+**Constraints:**
+
+  * `1 <= grid.length == n <= 400`
+  * `grid[i].length == n`
+  * `grid[i][j]` is either `0` or `1`.
+  * There is at least one thief in the `grid`.
diff --git a/problems/2812-find-the-safest-path-in-a-grid/solution.go b/problems/2812-find-the-safest-path-in-a-grid/solution.go
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+package main
+
+import "container/heap"
+
+// maximumSafenessFactor returns the maximum safeness factor of any path from
+// (0,0) to (n-1,n-1), where the safeness factor of a path is the minimum
+// Manhattan distance from any cell on the path to the nearest thief.
+//
+// Approach (two phases):
+//  1. Multi-source BFS from every thief cell computes, for each cell, its
+//     Manhattan distance to the nearest thief. In a 4-connected grid the BFS
+//     step count equals that Manhattan distance.
+//  2. A Dijkstra-style walk with a max-heap finds the "widest" path: we always
+//     expand the reachable cell with the largest bottleneck value, where a
+//     path's value is the minimum dist over its cells. The first time we pop
+//     the destination, its recorded value is the answer.
+func maximumSafenessFactor(grid [][]int) int {
+	n := len(grid)
+	if n == 0 {
+		return 0
+	}
+
+	// Phase 1: distance to nearest thief via multi-source BFS.
+	dist := make([][]int, n)
+	for i := range dist {
+		dist[i] = make([]int, n)
+		for j := range dist[i] {
+			dist[i][j] = -1
+		}
+	}
+
+	type cell struct{ r, c int }
+	queue := make([]cell, 0, n*n)
+	for r := 0; r < n; r++ {
+		for c := 0; c < n; c++ {
+			if grid[r][c] == 1 {
+				dist[r][c] = 0
+				queue = append(queue, cell{r, c})
+			}
+		}
+	}
+
+	dirs := [4][2]int{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}
+	for len(queue) > 0 {
+		cur := queue[0]
+		queue = queue[1:]
+		for _, d := range dirs {
+			nr, nc := cur.r+d[0], cur.c+d[1]
+			if nr >= 0 && nr < n && nc >= 0 && nc < n && dist[nr][nc] == -1 {
+				dist[nr][nc] = dist[cur.r][cur.c] + 1
+				queue = append(queue, cell{nr, nc})
+			}
+		}
+	}
+
+	// Phase 2: widest path via a max-heap keyed by path safeness.
+	safe := make([][]int, n)
+	for i := range safe {
+		safe[i] = make([]int, n)
+		for j := range safe[i] {
+			safe[i][j] = -1
+		}
+	}
+
+	pq := &maxHeap{{dist[0][0], 0, 0}}
+	heap.Init(pq)
+	safe[0][0] = dist[0][0]
+
+	for pq.Len() > 0 {
+		item := heap.Pop(pq).(hnode)
+		if item.r == n-1 && item.c == n-1 {
+			return item.val
+		}
+		// Skip stale heap entries superseded by a better path.
+		if item.val < safe[item.r][item.c] {
+			continue
+		}
+		for _, d := range dirs {
+			nr, nc := item.r+d[0], item.c+d[1]
+			if nr < 0 || nr >= n || nc < 0 || nc >= n {
+				continue
+			}
+			nv := item.val
+			if dist[nr][nc] < nv {
+				nv = dist[nr][nc]
+			}
+			if nv > safe[nr][nc] {
+				safe[nr][nc] = nv
+				heap.Push(pq, hnode{nv, nr, nc})
+			}
+		}
+	}
+
+	return 0
+}
+
+// hnode is a heap entry: val is the path safeness to reach cell (r, c).
+type hnode struct{ val, r, c int }
+
+// maxHeap orders entries so the largest safeness is popped first.
+type maxHeap []hnode
+
+func (h maxHeap) Len() int           { return len(h) }
+func (h maxHeap) Less(i, j int) bool { return h[i].val > h[j].val }
+func (h maxHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
+
+func (h *maxHeap) Push(x any) { *h = append(*h, x.(hnode)) }
+
+func (h *maxHeap) Pop() any {
+	old := *h
+	n := len(old)
+	item := old[n-1]
+	*h = old[:n-1]
+	return item
+}
diff --git a/problems/2812-find-the-safest-path-in-a-grid/solution_test.go b/problems/2812-find-the-safest-path-in-a-grid/solution_test.go
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+package main
+
+import "testing"
+
+func TestSolution(t *testing.T) {
+	tests := []struct {
+		name     string
+		grid     [][]int
+		expected int
+	}{
+		{
+			name:     "example 1: thieves block both corners",
+			grid:     [][]int{{1, 0, 0}, {0, 0, 0}, {0, 0, 1}},
+			expected: 0,
+		},
+		{
+			name:     "example 2: single thief top-right",
+			grid:     [][]int{{0, 0, 1}, {0, 0, 0}, {0, 0, 0}},
+			expected: 2,
+		},
+		{
+			name:     "example 3: thieves on opposite corners of 4x4",
+			grid:     [][]int{{0, 0, 0, 1}, {0, 0, 0, 0}, {0, 0, 0, 0}, {1, 0, 0, 0}},
+			expected: 2,
+		},
+		{
+			name:     "edge case: single cell that is a thief",
+			grid:     [][]int{{1}},
+			expected: 0,
+		},
+		{
+			name:     "edge case: every cell is a thief",
+			grid:     [][]int{{1, 1}, {1, 1}},
+			expected: 0,
+		},
+		{
+			name:     "edge case: start cell is a thief",
+			grid:     [][]int{{1, 0, 0}, {0, 0, 0}, {0, 0, 0}},
+			expected: 0,
+		},
+		{
+			name:     "edge case: single thief in the center of 5x5",
+			grid:     [][]int{{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}},
+			expected: 2,
+		},
+	}
+
+	for _, tt := range tests {
+		t.Run(tt.name, func(t *testing.T) {
+			result := maximumSafenessFactor(tt.grid)
+			if result != tt.expected {
+				t.Errorf("maximumSafenessFactor(%v) = %d, want %d", tt.grid, result, tt.expected)
+			}
+		})
+	}
+}