How do you solve As Far from Land as Possible on LeetCode?

As Far from Land as Possible (LeetCode #1162) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Using BFS allows us to explore all paths simultaneously, which is more efficient than checking each water cell individually.

What is the time complexity of As Far from Land as Possible?

The optimal solution for As Far from Land as Possible runs in O(n²) time and uses O(n) space. The time complexity is O(n²) due to visiting each cell once, while the space complexity is O(n) for the queue used in BFS.

What is the brute force solution for As Far from Land as Possible?

The brute force approach for As Far from Land as Possible is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check each water cell and calculate its distance to the nearest land cell. This is straightforward but inefficient, as we will repeatedly calculate distances for each water cell. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve As Far from Land as Possible?

As Far from Land as Possible uses the following concepts: Array, Dynamic Programming, Breadth-First Search, Matrix. The recommended patterns to study are: Breadth-First Search, Matrix Traversal.

What are the interview tips for As Far from Land as Possible?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases and how they might affect your solution. Explain your thought process clearly during the interview; it shows your understanding of the problem.

What are common mistakes when solving As Far from Land as Possible?

Forgetting to check edge cases, such as when there are no land or water cells. Not initializing the BFS queue correctly or handling visited cells properly.

#1162

As Far from Land as Possible

Medium

ArrayDynamic ProgrammingBreadth-First SearchMatrixBreadth-First SearchMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

Instead of checking each water cell, we can use a Breadth-First Search (BFS) starting from all land cells simultaneously. This way, we can efficiently calculate the distance to the nearest land for all water cells in one pass.

⚙️

Algorithm

3 steps

1Step 1: Initialize a queue and add all land cells to it.
2Step 2: Perform BFS from all land cells, marking distances for each water cell as we go.
3Step 3: Track the maximum distance encountered during the BFS.

solution.py24 lines

1from collections import deque
2
3def maxDistance(grid):
4    n = len(grid)
5    queue = deque()
6    for i in range(n):
7        for j in range(n):
8            if grid[i][j] == 1:
9                queue.append((i, j))
10    if not queue or len(queue) == n * n:
11        return -1
12
13    directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
14    max_dist = -1
15    while queue:
16        for _ in range(len(queue)):
17            x, y = queue.popleft()
18            for dx, dy in directions:
19                nx, ny = x + dx, y + dy
20                if 0 <= nx < n and 0 <= ny < n and grid[nx][ny] == 0:
21                    grid[nx][ny] = grid[x][y] + 1
22                    max_dist = max(max_dist, grid[nx][ny])
23                    queue.append((nx, ny))
24    return max_dist - 1

ℹ

Complexity note: The time complexity is O(n²) due to visiting each cell once, while the space complexity is O(n) for the queue used in BFS.

1Using BFS allows us to explore all paths simultaneously, which is more efficient than checking each water cell individually.
2Manhattan distance is straightforward to calculate, but understanding how to optimize the search is key.

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