LeetCode #2258 — HARD

Escape the Spreading Fire

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a 0-indexed 2D integer array grid of size m x n which represents a field. Each cell has one of three values:

0 represents grass,
1 represents fire,
2 represents a wall that you and fire cannot pass through.

You are situated in the top-left cell, (0, 0), and you want to travel to the safehouse at the bottom-right cell, (m - 1, n - 1). Every minute, you may move to an adjacent grass cell. After your move, every fire cell will spread to all adjacent cells that are not walls.

Return the maximum number of minutes that you can stay in your initial position before moving while still safely reaching the safehouse. If this is impossible, return -1. If you can always reach the safehouse regardless of the minutes stayed, return 10⁹.

Note that even if the fire spreads to the safehouse immediately after you have reached it, it will be counted as safely reaching the safehouse.

A cell is adjacent to another cell if the former is directly north, east, south, or west of the latter (i.e., their sides are touching).

Example 1:

Input: grid = [[0,2,0,0,0,0,0],[0,0,0,2,2,1,0],[0,2,0,0,1,2,0],[0,0,2,2,2,0,2],[0,0,0,0,0,0,0]]
Output: 3
Explanation: The figure above shows the scenario where you stay in the initial position for 3 minutes.
You will still be able to safely reach the safehouse.
Staying for more than 3 minutes will not allow you to safely reach the safehouse.

Example 2:

Input: grid = [[0,0,0,0],[0,1,2,0],[0,2,0,0]]
Output: -1
Explanation: The figure above shows the scenario where you immediately move towards the safehouse.
Fire will spread to any cell you move towards and it is impossible to safely reach the safehouse.
Thus, -1 is returned.

Example 3:

Input: grid = [[0,0,0],[2,2,0],[1,2,0]]
Output: 1000000000
Explanation: The figure above shows the initial grid.
Notice that the fire is contained by walls and you will always be able to safely reach the safehouse.
Thus, 10⁹ is returned.

Constraints:

m == grid.length
n == grid[i].length
2 <= m, n <= 300
4 <= m * n <= 2 * 10⁴
grid[i][j] is either 0, 1, or 2.
grid[0][0] == grid[m - 1][n - 1] == 0

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed 2D integer array grid of size m x n which represents a field. Each cell has one of three values: 0 represents grass, 1 represents fire, 2 represents a wall that you and fire cannot pass through. You are situated in the top-left cell, (0, 0), and you want to travel to the safehouse at the bottom-right cell, (m - 1, n - 1). Every minute, you may move to an adjacent grass cell. After your move, every fire cell will spread to all adjacent cells that are not walls. Return the maximum number of minutes that you can stay in your initial position before moving while still safely reaching the safehouse. If this is impossible, return -1. If you can always reach the safehouse regardless of the minutes stayed, return 109. Note that even if the fire spreads to the safehouse immediately after you have reached it, it will be counted as safely reaching the safehouse. A cell is

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

[[0,2,0,0,0,0,0],[0,0,0,2,2,1,0],[0,2,0,0,1,2,0],[0,0,2,2,2,0,2],[0,0,0,0,0,0,0]]

Example 2

[[0,0,0,0],[0,1,2,0],[0,2,0,0]]

Example 3

[[0,0,0],[2,2,0],[1,2,0]]

Core Insight

What unlocks the optimal approach

For some tile (x, y), how can we determine when, if ever, the fire will reach it?
We can use multi-source BFS to find the earliest time the fire will reach each cell.
Then, starting with a given t minutes of staying in the initial position, we can check if there is a safe path to the safehouse using the obtained information about the fire.
We can use binary search to efficiently find the maximum t that allows us to reach the safehouse.

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

Largest constraint values

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 #2258: Escape the Spreading Fire
class Solution {
    private int[][] grid;
    private boolean[][] fire;
    private boolean[][] vis;
    private final int[] dirs = {-1, 0, 1, 0, -1};
    private int m;
    private int n;

    public int maximumMinutes(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        fire = new boolean[m][n];
        vis = new boolean[m][n];
        int l = -1, r = m * n;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (check(mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l == m * n ? 1000000000 : l;
    }

    private boolean check(int t) {
        for (int i = 0; i < m; ++i) {
            Arrays.fill(fire[i], false);
            Arrays.fill(vis[i], false);
        }
        Deque<int[]> q1 = new ArrayDeque<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    q1.offer(new int[] {i, j});
                    fire[i][j] = true;
                }
            }
        }
        for (; t > 0 && !q1.isEmpty(); --t) {
            q1 = spread(q1);
        }
        if (fire[0][0]) {
            return false;
        }
        Deque<int[]> q2 = new ArrayDeque<>();
        q2.offer(new int[] {0, 0});
        vis[0][0] = true;
        for (; !q2.isEmpty(); q1 = spread(q1)) {
            for (int d = q2.size(); d > 0; --d) {
                int[] p = q2.poll();
                if (fire[p[0]][p[1]]) {
                    continue;
                }
                for (int k = 0; k < 4; ++k) {
                    int x = p[0] + dirs[k], y = p[1] + dirs[k + 1];
                    if (x >= 0 && x < m && y >= 0 && y < n && !fire[x][y] && !vis[x][y]
                        && grid[x][y] == 0) {
                        if (x == m - 1 && y == n - 1) {
                            return true;
                        }
                        vis[x][y] = true;
                        q2.offer(new int[] {x, y});
                    }
                }
            }
        }
        return false;
    }

    private Deque<int[]> spread(Deque<int[]> q) {
        Deque<int[]> nq = new ArrayDeque<>();
        while (!q.isEmpty()) {
            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 < m && y >= 0 && y < n && !fire[x][y] && grid[x][y] == 0) {
                    fire[x][y] = true;
                    nq.offer(new int[] {x, y});
                }
            }
        }
        return nq;
    }
}

// Accepted solution for LeetCode #2258: Escape the Spreading Fire
func maximumMinutes(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	fire := make([][]bool, m)
	vis := make([][]bool, m)
	dirs := [5]int{-1, 0, 1, 0, -1}
	for i := range fire {
		fire[i] = make([]bool, n)
		vis[i] = make([]bool, n)
	}
	l, r := -1, m*n
	spread := func(q [][2]int) [][2]int {
		nq := [][2]int{}
		for len(q) > 0 {
			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 < m && y >= 0 && y < n && !fire[x][y] && grid[x][y] == 0 {
					fire[x][y] = true
					nq = append(nq, [2]int{x, y})
				}
			}
		}
		return nq
	}
	check := func(t int) bool {
		for i := range fire {
			for j := range fire[i] {
				fire[i][j] = false
				vis[i][j] = false
			}
		}
		q1 := [][2]int{}
		for i := 0; i < m; i++ {
			for j := 0; j < n; j++ {
				if grid[i][j] == 1 {
					q1 = append(q1, [2]int{i, j})
					fire[i][j] = true
				}
			}
		}
		for ; t > 0 && len(q1) > 0; t-- {
			q1 = spread(q1)
		}
		q2 := [][2]int{{0, 0}}
		vis[0][0] = true
		for ; len(q2) > 0; q1 = spread(q1) {
			for d := len(q2); d > 0; d-- {
				p := q2[0]
				q2 = q2[1:]
				if fire[p[0]][p[1]] {
					continue
				}
				for k := 0; k < 4; k++ {
					x, y := p[0]+dirs[k], p[1]+dirs[k+1]
					if x >= 0 && x < m && y >= 0 && y < n && !fire[x][y] && !vis[x][y] && grid[x][y] == 0 {
						if x == m-1 && y == n-1 {
							return true
						}
						vis[x][y] = true
						q2 = append(q2, [2]int{x, y})
					}
				}
			}
		}
		return false
	}
	for l < r {
		mid := (l + r + 1) >> 1
		if check(mid) {
			l = mid
		} else {
			r = mid - 1
		}
	}
	if l == m*n {
		return int(1e9)
	}
	return l
}

# Accepted solution for LeetCode #2258: Escape the Spreading Fire
class Solution:
    def maximumMinutes(self, grid: List[List[int]]) -> int:
        def spread(q: Deque[int]) -> Deque[int]:
            nq = deque()
            while q:
                i, j = q.popleft()
                for a, b in pairwise(dirs):
                    x, y = i + a, j + b
                    if 0 <= x < m and 0 <= y < n and not fire[x][y] and grid[x][y] == 0:
                        fire[x][y] = True
                        nq.append((x, y))
            return nq

        def check(t: int) -> bool:
            for i in range(m):
                for j in range(n):
                    fire[i][j] = False
            q1 = deque()
            for i, row in enumerate(grid):
                for j, x in enumerate(row):
                    if x == 1:
                        fire[i][j] = True
                        q1.append((i, j))
            while t and q1:
                q1 = spread(q1)
                t -= 1
            if fire[0][0]:
                return False
            q2 = deque([(0, 0)])
            vis = [[False] * n for _ in range(m)]
            vis[0][0] = True
            while q2:
                for _ in range(len(q2)):
                    i, j = q2.popleft()
                    if fire[i][j]:
                        continue
                    for a, b in pairwise(dirs):
                        x, y = i + a, j + b
                        if (
                            0 <= x < m
                            and 0 <= y < n
                            and not vis[x][y]
                            and not fire[x][y]
                            and grid[x][y] == 0
                        ):
                            if x == m - 1 and y == n - 1:
                                return True
                            vis[x][y] = True
                            q2.append((x, y))
                q1 = spread(q1)
            return False

        m, n = len(grid), len(grid[0])
        l, r = -1, m * n
        dirs = (-1, 0, 1, 0, -1)
        fire = [[False] * n for _ in range(m)]
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return int(1e9) if l == m * n else l

// Accepted solution for LeetCode #2258: Escape the Spreading Fire
/**
 * [2258] Escape the Spreading Fire
 *
 * You are given a 0-indexed 2D integer array grid of size m x n which represents a field. Each cell has one of three values:
 *
 * 	0 represents grass,
 * 	1 represents fire,
 * 	2 represents a wall that you and fire cannot pass through.
 *
 * You are situated in the top-left cell, (0, 0), and you want to travel to the safehouse at the bottom-right cell, (m - 1, n - 1). Every minute, you may move to an adjacent grass cell. After your move, every fire cell will spread to all adjacent cells that are not walls.
 * Return the maximum number of minutes that you can stay in your initial position before moving while still safely reaching the safehouse. If this is impossible, return -1. If you can always reach the safehouse regardless of the minutes stayed, return 10^9.
 * Note that even if the fire spreads to the safehouse immediately after you have reached it, it will be counted as safely reaching the safehouse.
 * A cell is adjacent to another cell if the former is directly north, east, south, or west of the latter (i.e., their sides are touching).
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2022/03/10/ex1new.jpg" style="width: 650px; height: 404px;" />
 * Input: grid = [[0,2,0,0,0,0,0],[0,0,0,2,2,1,0],[0,2,0,0,1,2,0],[0,0,2,2,2,0,2],[0,0,0,0,0,0,0]]
 * Output: 3
 * Explanation: The figure above shows the scenario where you stay in the initial position for 3 minutes.
 * You will still be able to safely reach the safehouse.
 * Staying for more than 3 minutes will not allow you to safely reach the safehouse.
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2022/03/10/ex2new2.jpg" style="width: 515px; height: 150px;" />
 * Input: grid = [[0,0,0,0],[0,1,2,0],[0,2,0,0]]
 * Output: -1
 * Explanation: The figure above shows the scenario where you immediately move towards the safehouse.
 * Fire will spread to any cell you move towards and it is impossible to safely reach the safehouse.
 * Thus, -1 is returned.
 *
 * Example 3:
 * <img alt="" src="https://assets.leetcode.com/uploads/2022/03/10/ex3new.jpg" style="width: 174px; height: 150px;" />
 * Input: grid = [[0,0,0],[2,2,0],[1,2,0]]
 * Output: 1000000000
 * Explanation: The figure above shows the initial grid.
 * Notice that the fire is contained by walls and you will always be able to safely reach the safehouse.
 * Thus, 10^9 is returned.
 *
 *  
 * Constraints:
 *
 * 	m == grid.length
 * 	n == grid[i].length
 * 	2 <= m, n <= 300
 * 	4 <= m * n <= 2 * 10^4
 * 	grid[i][j] is either 0, 1, or 2.
 * 	grid[0][0] == grid[m - 1][n - 1] == 0
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/escape-the-spreading-fire/
// discuss: https://leetcode.com/problems/escape-the-spreading-fire/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn maximum_minutes(grid: Vec<Vec<i32>>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2258_example_1() {
        let grid = vec![
            vec![0, 2, 0, 0, 0, 0, 0],
            vec![0, 0, 0, 2, 2, 1, 0],
            vec![0, 2, 0, 0, 1, 2, 0],
            vec![0, 0, 2, 2, 2, 0, 2],
            vec![0, 0, 0, 0, 0, 0, 0],
        ];

        let result = 3;

        assert_eq!(Solution::maximum_minutes(grid), result);
    }

    #[test]
    #[ignore]
    fn test_2258_example_2() {
        let grid = vec![vec![0, 0, 0, 0], vec![0, 1, 2, 0], vec![0, 2, 0, 0]];

        let result = -1;

        assert_eq!(Solution::maximum_minutes(grid), result);
    }

    #[test]
    #[ignore]
    fn test_2258_example_3() {
        let grid = vec![vec![0, 0, 0], vec![2, 2, 0], vec![1, 2, 0]];

        let result = 1000000000;

        assert_eq!(Solution::maximum_minutes(grid), result);
    }
}

// Accepted solution for LeetCode #2258: Escape the Spreading Fire
function maximumMinutes(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    const fire = Array.from({ length: m }, () => Array.from({ length: n }, () => false));
    const vis = Array.from({ length: m }, () => Array.from({ length: n }, () => false));
    const dirs: number[] = [-1, 0, 1, 0, -1];
    let [l, r] = [-1, m * n];
    const spread = (q: number[][]): number[][] => {
        const nq: number[][] = [];
        while (q.length) {
            const [i, j] = q.shift()!;
            for (let k = 0; k < 4; ++k) {
                const [x, y] = [i + dirs[k], j + dirs[k + 1]];
                if (x >= 0 && x < m && y >= 0 && y < n && !fire[x][y] && grid[x][y] === 0) {
                    fire[x][y] = true;
                    nq.push([x, y]);
                }
            }
        }
        return nq;
    };
    const check = (t: number): boolean => {
        for (let i = 0; i < m; ++i) {
            fire[i].fill(false);
            vis[i].fill(false);
        }
        let q1: number[][] = [];
        for (let i = 0; i < m; ++i) {
            for (let j = 0; j < n; ++j) {
                if (grid[i][j] === 1) {
                    q1.push([i, j]);
                    fire[i][j] = true;
                }
            }
        }
        for (; t && q1.length; --t) {
            q1 = spread(q1);
        }
        if (fire[0][0]) {
            return false;
        }
        const q2: number[][] = [[0, 0]];
        vis[0][0] = true;
        for (; q2.length; q1 = spread(q1)) {
            for (let d = q2.length; d; --d) {
                const [i, j] = q2.shift()!;
                if (fire[i][j]) {
                    continue;
                }
                for (let k = 0; k < 4; ++k) {
                    const [x, y] = [i + dirs[k], j + dirs[k + 1]];
                    if (
                        x >= 0 &&
                        x < m &&
                        y >= 0 &&
                        y < n &&
                        !vis[x][y] &&
                        !fire[x][y] &&
                        grid[x][y] === 0
                    ) {
                        if (x === m - 1 && y === n - 1) {
                            return true;
                        }
                        vis[x][y] = true;
                        q2.push([x, y]);
                    }
                }
            }
        }
        return false;
    };
    while (l < r) {
        const mid = (l + r + 1) >> 1;
        if (check(mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }
    return l === m * n ? 1e9 : l;
}

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(m × n × log (m × n)

Space

O(m × n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.