LeetCode #3341 — MEDIUM

Find Minimum Time to Reach Last Room I

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

The Problem

Problem Statement

There is a dungeon with n x m rooms arranged as a grid.

You are given a 2D array moveTime of size n x m, where moveTime[i][j] represents the minimum time in seconds after which the room opens and can be moved to. You start from the room (0, 0) at time t = 0 and can move to an adjacent room. Moving between adjacent rooms takes exactly one second.

Return the minimum time to reach the room (n - 1, m - 1).

Two rooms are adjacent if they share a common wall, either horizontally or vertically.

Example 1:

Input: moveTime = [[0,4],[4,4]]

Output: 6

Explanation:

The minimum time required is 6 seconds.

At time t == 4, move from room (0, 0) to room (1, 0) in one second.
At time t == 5, move from room (1, 0) to room (1, 1) in one second.

Example 2:

Input: moveTime = [[0,0,0],[0,0,0]]

Output: 3

Explanation:

The minimum time required is 3 seconds.

At time t == 0, move from room (0, 0) to room (1, 0) in one second.
At time t == 1, move from room (1, 0) to room (1, 1) in one second.
At time t == 2, move from room (1, 1) to room (1, 2) in one second.

Example 3:

Input: moveTime = [[0,1],[1,2]]

Output: 3

Constraints:

2 <= n == moveTime.length <= 50
2 <= m == moveTime[i].length <= 50
0 <= moveTime[i][j] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: There is a dungeon with n x m rooms arranged as a grid. You are given a 2D array moveTime of size n x m, where moveTime[i][j] represents the minimum time in seconds after which the room opens and can be moved to. You start from the room (0, 0) at time t = 0 and can move to an adjacent room. Moving between adjacent rooms takes exactly one second. Return the minimum time to reach the room (n - 1, m - 1). Two rooms are adjacent if they share a common wall, either horizontally or vertically.

Baseline thinking

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

Pattern signal: Array

Example 1

[[0,4],[4,4]]

Example 2

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

Example 3

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

Core Insight

What unlocks the optimal approach

Use shortest path algorithms.

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 #3341: Find Minimum Time to Reach Last Room I
class Solution {
    public int minTimeToReach(int[][] moveTime) {
        int n = moveTime.length;
        int m = moveTime[0].length;
        int[][] dist = new int[n][m];
        for (var row : dist) {
            Arrays.fill(row, Integer.MAX_VALUE);
        }
        dist[0][0] = 0;

        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        pq.offer(new int[] {0, 0, 0});
        int[] dirs = {-1, 0, 1, 0, -1};
        while (true) {
            int[] p = pq.poll();
            int d = p[0], i = p[1], j = p[2];

            if (i == n - 1 && j == m - 1) {
                return d;
            }
            if (d > dist[i][j]) {
                continue;
            }

            for (int k = 0; k < 4; k++) {
                int x = i + dirs[k];
                int y = j + dirs[k + 1];
                if (x >= 0 && x < n && y >= 0 && y < m) {
                    int t = Math.max(moveTime[x][y], dist[i][j]) + 1;
                    if (dist[x][y] > t) {
                        dist[x][y] = t;
                        pq.offer(new int[] {t, x, y});
                    }
                }
            }
        }
    }
}

// Accepted solution for LeetCode #3341: Find Minimum Time to Reach Last Room I
func minTimeToReach(moveTime [][]int) int {
	n, m := len(moveTime), len(moveTime[0])
	dist := make([][]int, n)
	for i := range dist {
		dist[i] = make([]int, m)
		for j := range dist[i] {
			dist[i][j] = math.MaxInt32
		}
	}
	dist[0][0] = 0

	pq := &hp{}
	heap.Init(pq)
	heap.Push(pq, tuple{0, 0, 0})

	dirs := []int{-1, 0, 1, 0, -1}
	for {
		p := heap.Pop(pq).(tuple)
		d, i, j := p.dis, p.x, p.y

		if i == n-1 && j == m-1 {
			return d
		}
		if d > dist[i][j] {
			continue
		}

		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < n && y >= 0 && y < m {
				t := max(moveTime[x][y], dist[i][j]) + 1
				if dist[x][y] > t {
					dist[x][y] = t
					heap.Push(pq, tuple{t, x, y})
				}
			}
		}
	}
}

type tuple struct{ dis, x, y int }
type hp []tuple

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].dis < h[j].dis }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any)        { *h = append(*h, v.(tuple)) }
func (h *hp) Pop() (v any)      { a := *h; *h, v = a[:len(a)-1], a[len(a)-1]; return }

# Accepted solution for LeetCode #3341: Find Minimum Time to Reach Last Room I
class Solution:
    def minTimeToReach(self, moveTime: List[List[int]]) -> int:
        n, m = len(moveTime), len(moveTime[0])
        dist = [[inf] * m for _ in range(n)]
        dist[0][0] = 0
        pq = [(0, 0, 0)]
        dirs = (-1, 0, 1, 0, -1)
        while 1:
            d, i, j = heappop(pq)
            if i == n - 1 and j == m - 1:
                return d
            if d > dist[i][j]:
                continue
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < n and 0 <= y < m:
                    t = max(moveTime[x][y], dist[i][j]) + 1
                    if dist[x][y] > t:
                        dist[x][y] = t
                        heappush(pq, (t, x, y))

// Accepted solution for LeetCode #3341: Find Minimum Time to Reach Last Room I
/**
 * [3341] Find Minimum Time to Reach Last Room I
 */
pub struct Solution {}

// submission codes start here
use std::cmp::Ordering;
use std::collections::BinaryHeap;

const DIRECTIONS: [(isize, isize); 4] = [(0, 1), (1, 0), (-1, 0), (0, -1)];

#[derive(Debug, Copy, Clone, Eq, PartialEq, Ord)]
struct State {
    x: usize,
    y: usize,
    distance: i32,
}

impl State {
    fn new(x: usize, y: usize, distance: i32) -> Self {
        Self { x, y, distance }
    }
}

impl PartialOrd for State {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        // BinaryHeap是大顶堆
        Some(other.distance.cmp(&self.distance))
    }
}

impl Solution {
    pub fn min_time_to_reach(move_time: Vec<Vec<i32>>) -> i32 {
        let n = move_time.len();
        let m = move_time[0].len();

        let mut distances = vec![vec![i32::MAX; m]; n];
        let mut visited = vec![vec![false; m]; n];
        let mut heap = BinaryHeap::new();

        distances[0][0] = 0;
        heap.push(State::new(0, 0, 0));

        while let Some(head) = heap.pop() {
            if visited[head.x][head.y] {
                continue;
            }
            visited[head.x][head.y] = true;

            for &(dx, dy) in DIRECTIONS.iter() {
                let next = head
                    .x
                    .checked_add_signed(dx)
                    .and_then(|x| head.y.checked_add_signed(dy).and_then(|y| Some((x, y))))
                    .filter(|(x, y)| x < &n && y < &m);

                if let Some((x, y)) = next {
                    let distance = distances[head.x][head.y].max(move_time[x][y]) + 1;

                    if distances[x][y] > distance {
                        distances[x][y] = distance;
                        heap.push(State::new(x, y, distance));
                    }
                }
            }
        }

        distances[n - 1][m - 1]
    }
}

// submission codes end

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

    #[test]
    fn test_3341() {
        assert_eq!(6, Solution::min_time_to_reach(vec![vec![0, 4], vec![4, 4]]));
        assert_eq!(
            3,
            Solution::min_time_to_reach(vec![vec![0, 0, 0], vec![0, 0, 0]])
        );
        assert_eq!(3, Solution::min_time_to_reach(vec![vec![0, 1], vec![1, 2]]));
    }
}

// Accepted solution for LeetCode #3341: Find Minimum Time to Reach Last Room I
function minTimeToReach(moveTime: number[][]): number {
    const n = moveTime.length;
    const m = moveTime[0].length;
    const dist = Array.from({ length: n }, () => Array(m).fill(Infinity));
    dist[0][0] = 0;
    type Node = [number, number, number];
    const pq = new PriorityQueue<Node>((a, b) => a[0] - b[0]);
    pq.enqueue([0, 0, 0]);
    const dirs = [-1, 0, 1, 0, -1];
    while (!pq.isEmpty()) {
        const [d, i, j] = pq.dequeue();
        if (d > dist[i][j]) continue;
        if (i === n - 1 && j === m - 1) return d;
        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 < m) {
                const t = Math.max(moveTime[x][y], d) + 1;
                if (t < dist[x][y]) {
                    dist[x][y] = t;
                    pq.enqueue([t, x, y]);
                }
            }
        }
    }
    return -1;
}

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 × m × log (n × m)

Space

O(n × m)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.