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.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given a m x n matrix grid consisting of non-negative integers where grid[row][col] represents the minimum time required to be able to visit the cell (row, col), which means you can visit the cell (row, col) only when the time you visit it is greater than or equal to grid[row][col].
You are standing in the top-left cell of the matrix in the 0th second, and you must move to any adjacent cell in the four directions: up, down, left, and right. Each move you make takes 1 second.
Return the minimum time required in which you can visit the bottom-right cell of the matrix. If you cannot visit the bottom-right cell, then return -1.
Example 1:
Input: grid = [[0,1,3,2],[5,1,2,5],[4,3,8,6]] Output: 7 Explanation: One of the paths that we can take is the following: - at t = 0, we are on the cell (0,0). - at t = 1, we move to the cell (0,1). It is possible because grid[0][1] <= 1. - at t = 2, we move to the cell (1,1). It is possible because grid[1][1] <= 2. - at t = 3, we move to the cell (1,2). It is possible because grid[1][2] <= 3. - at t = 4, we move to the cell (1,1). It is possible because grid[1][1] <= 4. - at t = 5, we move to the cell (1,2). It is possible because grid[1][2] <= 5. - at t = 6, we move to the cell (1,3). It is possible because grid[1][3] <= 6. - at t = 7, we move to the cell (2,3). It is possible because grid[2][3] <= 7. The final time is 7. It can be shown that it is the minimum time possible.
Example 2:
Input: grid = [[0,2,4],[3,2,1],[1,0,4]] Output: -1 Explanation: There is no path from the top left to the bottom-right cell.
Constraints:
m == grid.lengthn == grid[i].length2 <= m, n <= 10004 <= m * n <= 1050 <= grid[i][j] <= 105grid[0][0] == 0Problem summary: You are given a m x n matrix grid consisting of non-negative integers where grid[row][col] represents the minimum time required to be able to visit the cell (row, col), which means you can visit the cell (row, col) only when the time you visit it is greater than or equal to grid[row][col]. You are standing in the top-left cell of the matrix in the 0th second, and you must move to any adjacent cell in the four directions: up, down, left, and right. Each move you make takes 1 second. Return the minimum time required in which you can visit the bottom-right cell of the matrix. If you cannot visit the bottom-right cell, then return -1.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[[0,1,3,2],[5,1,2,5],[4,3,8,6]]
[[0,2,4],[3,2,1],[1,0,4]]
find-minimum-time-to-reach-last-room-i)find-minimum-time-to-reach-last-room-ii)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2577: Minimum Time to Visit a Cell In a Grid
class Solution {
public int minimumTime(int[][] grid) {
if (grid[0][1] > 1 && grid[1][0] > 1) {
return -1;
}
int m = grid.length, n = grid[0].length;
int[][] dist = new int[m][n];
for (var e : dist) {
Arrays.fill(e, 1 << 30);
}
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) {
var p = pq.poll();
int i = p[1], j = p[2];
if (i == m - 1 && j == n - 1) {
return p[0];
}
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n) {
int nt = p[0] + 1;
if (nt < grid[x][y]) {
nt = grid[x][y] + (grid[x][y] - nt) % 2;
}
if (nt < dist[x][y]) {
dist[x][y] = nt;
pq.offer(new int[] {nt, x, y});
}
}
}
}
}
}
// Accepted solution for LeetCode #2577: Minimum Time to Visit a Cell In a Grid
func minimumTime(grid [][]int) int {
if grid[0][1] > 1 && grid[1][0] > 1 {
return -1
}
m, n := len(grid), len(grid[0])
dist := make([][]int, m)
for i := range dist {
dist[i] = make([]int, n)
for j := range dist[i] {
dist[i][j] = 1 << 30
}
}
dist[0][0] = 0
pq := hp{}
heap.Push(&pq, tuple{0, 0, 0})
dirs := [5]int{-1, 0, 1, 0, -1}
for {
p := heap.Pop(&pq).(tuple)
i, j := p.i, p.j
if i == m-1 && j == n-1 {
return p.t
}
for k := 0; k < 4; k++ {
x, y := i+dirs[k], j+dirs[k+1]
if x >= 0 && x < m && y >= 0 && y < n {
nt := p.t + 1
if nt < grid[x][y] {
nt = grid[x][y] + (grid[x][y]-nt)%2
}
if nt < dist[x][y] {
dist[x][y] = nt
heap.Push(&pq, tuple{nt, x, y})
}
}
}
}
}
type tuple struct{ t, i, j int }
type hp []tuple
func (h hp) Len() int { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].t < h[j].t }
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() any { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
# Accepted solution for LeetCode #2577: Minimum Time to Visit a Cell In a Grid
class Solution:
def minimumTime(self, grid: List[List[int]]) -> int:
if grid[0][1] > 1 and grid[1][0] > 1:
return -1
m, n = len(grid), len(grid[0])
dist = [[inf] * n for _ in range(m)]
dist[0][0] = 0
q = [(0, 0, 0)]
dirs = (-1, 0, 1, 0, -1)
while 1:
t, i, j = heappop(q)
if i == m - 1 and j == n - 1:
return t
for a, b in pairwise(dirs):
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n:
nt = t + 1
if nt < grid[x][y]:
nt = grid[x][y] + (grid[x][y] - nt) % 2
if nt < dist[x][y]:
dist[x][y] = nt
heappush(q, (nt, x, y))
// Accepted solution for LeetCode #2577: Minimum Time to Visit a Cell In a Grid
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2577: Minimum Time to Visit a Cell In a Grid
// class Solution {
// public int minimumTime(int[][] grid) {
// if (grid[0][1] > 1 && grid[1][0] > 1) {
// return -1;
// }
// int m = grid.length, n = grid[0].length;
// int[][] dist = new int[m][n];
// for (var e : dist) {
// Arrays.fill(e, 1 << 30);
// }
// 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) {
// var p = pq.poll();
// int i = p[1], j = p[2];
// if (i == m - 1 && j == n - 1) {
// return p[0];
// }
// for (int k = 0; k < 4; ++k) {
// int x = i + dirs[k], y = j + dirs[k + 1];
// if (x >= 0 && x < m && y >= 0 && y < n) {
// int nt = p[0] + 1;
// if (nt < grid[x][y]) {
// nt = grid[x][y] + (grid[x][y] - nt) % 2;
// }
// if (nt < dist[x][y]) {
// dist[x][y] = nt;
// pq.offer(new int[] {nt, x, y});
// }
// }
// }
// }
// }
// }
// Accepted solution for LeetCode #2577: Minimum Time to Visit a Cell In a Grid
function minimumTime(grid: number[][]): number {
if (grid[0][1] > 1 && grid[1][0] > 1) return -1;
const [m, n] = [grid.length, grid[0].length];
const DIRS = [-1, 0, 1, 0, -1];
const q = new PriorityQueue<number[]>((a, b) => a[0] - b[0]);
const dist: number[][] = Array.from({ length: m }, () =>
new Array(n).fill(Number.POSITIVE_INFINITY),
);
dist[0][0] = 0;
q.enqueue([0, 0, 0]);
while (true) {
const [t, i, j] = q.dequeue();
if (i === m - 1 && j === n - 1) return t;
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) continue;
let nt = t + 1;
if (nt < grid[x][y]) {
nt = grid[x][y] + ((grid[x][y] - nt) % 2);
}
if (nt < dist[x][y]) {
dist[x][y] = nt;
q.enqueue([nt, x, y]);
}
}
}
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.