Added solution for lc 2812

dsa-solutions/lc-solutions/2800-2899/2812-Find-the-Safest-Path-in-a-Grid.md

@@ -0,0 +1,136 @@
+---
+id: maximum-safeness-factor-of-a-path
+title: Maximum Safeness Factor of a Path
+sidebar_label: Maximum Safeness Factor of a Path
+tags:
+  - Grid
+  - Pathfinding
+  - BFS
+  - Dijkstra
+description: "This is a solution to the Maximum Safeness Factor of a Path problem."
+---
+
+## Problem Description
+
+You are given a 0-indexed 2D matrix `grid` of size `n x n`, where:
+
+- `grid[r][c] = 1` represents a cell containing a thief.
+- `grid[r][c] = 0` represents an empty cell.
+
+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`.
+
+### Examples
+
+**Example 1:**
+
+```
+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:**
+
+```
+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:**
+
+```
+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
+
+- The number of nodes in the grid is in the range `[1, 400]`.
+- `grid[i][j]` is either 0 or 1.
+- There is at least one thief in the grid.
+
+### Approach
+
+To solve this problem, we will first use a multi-source Breadth-First Search (BFS) to calculate the minimum distance from each cell to the nearest thief. Then, we will use a modified Dijkstra's algorithm to find the path from `(0, 0)` to `(n - 1, n - 1)` that maximizes the minimum safeness factor.
+
+#### Code in Python
+
+```python
+import heapq
+from collections import deque
+
+class Solution:
+    def maximumSafenessFactor(self, grid):
+        n = len(grid)
+
+        # Step 1: Calculate distance to nearest thief for each cell using multi-source BFS
+        distance = [[float('inf')] * n for _ in range(n)]
+        q = deque()
+
+        # Initialize the BFS queue with all thieves' positions
+        for r in range(n):
+            for c in range(n):
+                if grid[r][c] == 1:
+                    q.append((r, c))
+                    distance[r][c] = 0
+
+        # Directions for moving up, down, left, and right
+        directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
+
+        while q:
+            r, c = q.popleft()
+            for dr, dc in directions:
+                nr, nc = r + dr, c + dc
+                if 0 <= nr < n and 0 <= nc < n and distance[nr][nc] == float('inf'):
+                    distance[nr][nc] = distance[r][c] + 1
+                    q.append((nr, nc))
+
+        # Step 2: Use modified Dijkstra's algorithm to find the path with the maximum safeness factor
+        max_heap = [(-distance[0][0], 0, 0)]  # Max-heap (negative distance to simulate max-heap in python)
+        visited = set()
+
+        while max_heap:
+            min_dist, r, c = heapq.heappop(max_heap)
+            min_dist = -min_dist  # Convert back to positive
+
+            if (r, c) == (n-1, n-1):
+                return min_dist
+
+            if (r, c) in visited:
+                continue
+            visited.add((r, c))
+
+            for dr, dc in directions:
+                nr, nc = r + dr, c + dc
+                if 0 <= nr < n and 0 <= nc < n and (nr, nc) not in visited:
+                    heapq.heappush(max_heap, (-min(distance[nr][nc], min_dist), nr, nc))
+
+        return 0  # Fallback, though problem guarantees a path
+
+# Example test cases
+sol = Solution()
+grid1 = [[1,0,0],[0,0,0],[0,0,1]]
+print(sol.maximumSafenessFactor(grid1))  # Output: 0
+
+grid2 = [[0,0,1],[0,0,0],[0,0,0]]
+print(sol.maximumSafenessFactor(grid2))  # Output: 2
+
+grid3 = [[0,0,0,1],[0,0,0,0],[0,0,0,0],[1,0,0,0]]
+print(sol.maximumSafenessFactor(grid3))  # Output: 2
+```
+```