LeetCode #1162 — MEDIUM

As Far from Land as Possible

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an n x n grid containing only values 0 and 1, where 0 represents water and 1 represents land, find a water cell such that its distance to the nearest land cell is maximized, and return the distance. If no land or water exists in the grid, return -1.

The distance used in this problem is the Manhattan distance: the distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1|.

Example 1:

Input: grid = [[1,0,1],[0,0,0],[1,0,1]]
Output: 2
Explanation: The cell (1, 1) is as far as possible from all the land with distance 2.

Example 2:

Input: grid = [[1,0,0],[0,0,0],[0,0,0]]
Output: 4
Explanation: The cell (2, 2) is as far as possible from all the land with distance 4.

Constraints:

n == grid.length
n == grid[i].length
1 <= n <= 100
grid[i][j] is 0 or 1

Step 01

Brute Force Baseline

Problem summary: Given an n x n grid containing only values 0 and 1, where 0 represents water and 1 represents land, find a water cell such that its distance to the nearest land cell is maximized, and return the distance. If no land or water exists in the grid, return -1. The distance used in this problem is the Manhattan distance: the distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1|.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[[1,0,1],[0,0,0],[1,0,1]]

Example 2

[[1,0,0],[0,0,0],[0,0,0]]

Core Insight

What unlocks the optimal approach

Can you think of this problem in a backwards way ?
Imagine expanding outward from each land cell. What kind of search does that ?
Use BFS starting from all land cells in the same time.
When do you reach the furthest water cell?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1162: As Far from Land as Possible
class Solution {
    public int maxDistance(int[][] grid) {
        int n = grid.length;
        Deque<int[]> q = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    q.offer(new int[] {i, j});
                }
            }
        }
        int ans = -1;
        if (q.isEmpty() || q.size() == n * n) {
            return ans;
        }
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!q.isEmpty()) {
            for (int i = q.size(); i > 0; --i) {
                int[] p = q.poll();
                for (int k = 0; k < 4; ++k) {
                    int x = p[0] + dirs[k], y = p[1] + dirs[k + 1];
                    if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 0) {
                        grid[x][y] = 1;
                        q.offer(new int[] {x, y});
                    }
                }
            }
            ++ans;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1162: As Far from Land as Possible
func maxDistance(grid [][]int) int {
	n := len(grid)
	q := [][2]int{}
	for i, row := range grid {
		for j, v := range row {
			if v == 1 {
				q = append(q, [2]int{i, j})
			}
		}
	}
	ans := -1
	if len(q) == 0 || len(q) == n*n {
		return ans
	}
	dirs := [5]int{-1, 0, 1, 0, -1}
	for len(q) > 0 {
		for i := len(q); i > 0; i-- {
			p := q[0]
			q = q[1:]
			for k := 0; k < 4; k++ {
				x, y := p[0]+dirs[k], p[1]+dirs[k+1]
				if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 0 {
					grid[x][y] = 1
					q = append(q, [2]int{x, y})
				}
			}
		}
		ans++
	}
	return ans
}

# Accepted solution for LeetCode #1162: As Far from Land as Possible
class Solution:
    def maxDistance(self, grid: List[List[int]]) -> int:
        n = len(grid)
        q = deque((i, j) for i in range(n) for j in range(n) if grid[i][j])
        ans = -1
        if len(q) in (0, n * n):
            return ans
        dirs = (-1, 0, 1, 0, -1)
        while q:
            for _ in range(len(q)):
                i, j = q.popleft()
                for a, b in pairwise(dirs):
                    x, y = i + a, j + b
                    if 0 <= x < n and 0 <= y < n and grid[x][y] == 0:
                        grid[x][y] = 1
                        q.append((x, y))
            ans += 1
        return ans

// Accepted solution for LeetCode #1162: As Far from Land as Possible
struct Solution;
use std::collections::VecDeque;

impl Solution {
    fn max_distance(mut grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let m = grid[0].len();
        let mut queue: VecDeque<(usize, usize, i32)> = VecDeque::new();
        for i in 0..n {
            for j in 0..m {
                if grid[i][j] == 1 {
                    queue.push_back((i, j, 0));
                }
            }
        }
        let mut res = -1;
        let offsets = [(1, 0), (0, 1), (-1, 0), (0, -1)];
        while let Some((i, j, d)) = queue.pop_front() {
            for offset in &offsets {
                let i = i as i32 + offset.0;
                let j = j as i32 + offset.1;
                if i >= 0 && j >= 0 && i < n as i32 && j < m as i32 {
                    let i = i as usize;
                    let j = j as usize;
                    if grid[i][j] == 0 {
                        grid[i][j] = 1;
                        res = res.max(d + 1);
                        queue.push_back((i, j, d + 1));
                    }
                }
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let grid = vec_vec_i32![[1, 0, 1], [0, 0, 0], [1, 0, 1]];
    let res = 2;
    assert_eq!(Solution::max_distance(grid), res);
    let grid = vec_vec_i32![[1, 0, 0], [0, 0, 0], [0, 0, 0]];
    let res = 4;
    assert_eq!(Solution::max_distance(grid), res);
}

// Accepted solution for LeetCode #1162: As Far from Land as Possible
function maxDistance(grid: number[][]): number {
    const n = grid.length;
    const q: [number, number][] = [];
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] === 1) {
                q.push([i, j]);
            }
        }
    }
    let ans = -1;
    if (q.length === 0 || q.length === n * n) {
        return ans;
    }
    const dirs: number[] = [-1, 0, 1, 0, -1];
    while (q.length > 0) {
        for (let m = q.length; m; --m) {
            const [i, j] = q.shift()!;
            for (let k = 0; k < 4; ++k) {
                const x = i + dirs[k];
                const y = j + dirs[k + 1];
                if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] === 0) {
                    grid[x][y] = 1;
                    q.push([x, y]);
                }
            }
        }
        ++ans;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.