> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Minimum Time to Visit a Cell In a Grid

> Tested Python solution for LeetCode 2577 with 22 pytest cases. Generate a practice environment with lcpy.

LeetCode 2577, [Hard](/catalog/hard). Topics: [Array](/catalog/topics/array), [Breadth-First Search](/catalog/topics/breadth-first-search), [Graph Theory](/catalog/topics/graph-theory), [Heap (Priority Queue)](/catalog/topics/heap-priority-queue), [Matrix](/catalog/topics/matrix), [Shortest Path](/catalog/topics/shortest-path). [View on LeetCode](https://leetcode.com/problems/minimum-time-to-visit-a-cell-in-a-grid/description/).

Generate this problem as a practice environment: tested reference solution, 22 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 2577   # by problem number
lcpy gen -s minimum_time_to_visit_a_cell_in_a_grid   # by problem name
```

## Problem

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`.

### Examples

![Example 1](https://assets.leetcode.com/uploads/2023/02/14/yetgriddrawio-8.png)

```
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](https://assets.leetcode.com/uploads/2023/02/14/yetgriddrawio-9.png)

```
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.length
* n == grid\[i].length
* 2 \<= m, n \<= 1000
* 4 \<= m \* n \<= 10^5
* 0 \<= grid\[i]\[j] \<= 10^5
* grid\[0]\[0] == 0

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/minimum_time_to_visit_a_cell_in_a_grid/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/minimum_time_to_visit_a_cell_in_a_grid/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
import heapq


class Solution:
    # Time: O(m * n * log(m * n))
    # Space: O(m * n)
    def minimum_time(self, grid: list[list[int]]) -> int:
        if grid[0][1] > 1 and grid[1][0] > 1:
            return -1

        rows, cols = len(grid), len(grid[0])
        unvisited = 10**18
        best = [[unvisited] * cols for _ in range(rows)]
        best[0][0] = 0
        heap: list[tuple[int, int, int]] = [(0, 0, 0)]

        while heap:
            time, row, col = heapq.heappop(heap)
            if best[row][col] < time:
                continue
            if row == rows - 1 and col == cols - 1:
                return time

            for d_row, d_col in ((1, 0), (-1, 0), (0, 1), (0, -1)):
                n_row, n_col = row + d_row, col + d_col
                if not (0 <= n_row < rows and 0 <= n_col < cols):
                    continue
                need = grid[n_row][n_col]
                # Waiting means bouncing between two adjacent cells, which costs
                # 2 seconds per bounce, so the arrival parity is preserved.
                n_time = max(time + 1, need + ((need - time - 1) % 2))
                if n_time < best[n_row][n_col]:
                    best[n_row][n_col] = n_time
                    heapq.heappush(heap, (n_time, n_row, n_col))

        return -1
```

## Complexity

| Time | Space |
| - | - |
| O(m \* n \* log(m \* n)) | O(m \* n) |

## Tags

[NeetCode All](/catalog/neetcode).


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