How do you solve Find the Safest Path in a Grid on LeetCode?

Find the Safest Path in a Grid (LeetCode #2812) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² log n) time complexity and O(n²) space complexity. Key insight: Using BFS to calculate distances allows us to efficiently determine the safeness factor for paths.

What is the brute force solution for Find the Safest Path in a Grid?

The brute force approach for Find the Safest Path in a Grid is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach explores all possible paths from the starting cell to the destination cell, calculating the safeness factor for each path. This method is simple but inefficient, as it doesn't optimize for the best path. The optimal Optimal Solution approach improves this to O(n² log n).

What are the interview tips for Find the Safest Path in a Grid?

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches. Practice BFS and binary search problems to build confidence.

What are common mistakes when solving Find the Safest Path in a Grid?

Not considering all adjacent cells when performing BFS. Failing to handle edge cases, such as grids with no thieves.

#2812

Find the Safest Path in a Grid

Medium

ArrayBinary SearchBreadth-First SearchUnion-FindHeap (Priority Queue)MatrixBreadth-First SearchBinary Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n² log n)
Space	O(n)	O(n²)

💡

Intuition

Time O(n² log n)Space O(n²)

The optimal solution combines BFS to calculate the minimum distance from each cell to the nearest thief and uses binary search to find the maximum safeness factor efficiently. This approach avoids exploring all paths explicitly.

⚙️

Algorithm

3 steps

1Step 1: Perform a multi-source BFS starting from all cells containing thieves to calculate the minimum distance to each cell in the grid.
2Step 2: Use binary search to find the maximum safeness factor that allows a valid path from (0, 0) to (n-1, n-1).
3Step 3: For each mid value in binary search, check if there's a valid path using BFS that maintains at least that safeness factor.

solution.py47 lines

1from collections import deque
2
3def maxSafenessFactor(grid):
4    n = len(grid)
5    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
6    distance = [[float('inf')] * n for _ in range(n)]
7    queue = deque()
8
9    for i in range(n):
10        for j in range(n):
11            if grid[i][j] == 1:
12                queue.append((i, j))
13                distance[i][j] = 0
14
15    while queue:
16        x, y = queue.popleft()
17        for dx, dy in directions:
18            nx, ny = x + dx, y + dy
19            if 0 <= nx < n and 0 <= ny < n and distance[nx][ny] == float('inf'):
20                distance[nx][ny] = distance[x][y] + 1
21                queue.append((nx, ny))
22
23    def canReach(safeness):
24        visited = set()
25        queue = deque([(0, 0)])
26        visited.add((0, 0))
27        while queue:
28            x, y = queue.popleft()
29            if x == n - 1 and y == n - 1:
30                return True
31            for dx, dy in directions:
32                nx, ny = x + dx, y + dy
33                if 0 <= nx < n and 0 <= ny < n and (nx, ny) not in visited:
34                    if distance[nx][ny] >= safeness:
35                        visited.add((nx, ny))
36                        queue.append((nx, ny))
37        return False
38
39    left, right = 0, n * 2
40    while left < right:
41        mid = (left + right + 1) // 2
42        if canReach(mid):
43            left = mid
44        else:
45            right = mid - 1
46
47    return left

ℹ

Complexity note: The BFS runs in O(n²) to compute distances, and the binary search runs log(n²) times, leading to an overall complexity of O(n² log n). The space complexity is O(n²) due to the distance matrix.

1Using BFS to calculate distances allows us to efficiently determine the safeness factor for paths.
2Binary search helps in optimizing the search for the maximum safeness factor.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.