> ## 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.

# Swim in Rising Water Python Solution

> Tested Python solution for LeetCode 778 with 13 pytest cases. Generate a practice environment with lcpy.

LeetCode 778, Hard. Topics: Array, Binary Search, Depth-First Search, Breadth-First Search, Union-Find, Heap (Priority Queue), Matrix. [View on LeetCode](https://leetcode.com/problems/swim-in-rising-water/description/).

Generate this problem as a practice environment: tested reference solution, 13 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 778   # by problem number
lcpy gen -s swim_in_rising_water   # by problem name
```

## Problem

You are given an `n x n` integer matrix `grid` where each value `grid[i][j]` represents the elevation at that point `(i, j)`.

It starts raining, and water gradually rises over time. At time `t`, the water level is `t`, meaning **any** cell with elevation less than equal to `t` is submerged or reachable.

You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most `t`. You can swim infinite distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.

Return *the minimum time until you can reach the bottom right square* `(n - 1, n - 1)` *if you start at the top left square* `(0, 0)`.

### Examples

![Example 1](https://assets.leetcode.com/uploads/2021/06/29/swim1-grid.jpg)

```
Input: grid = [[0,2],[1,3]]
Output: 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.
```

![Example 2](https://assets.leetcode.com/uploads/2021/06/29/swim2-grid-1.jpg)

```
Input: grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
Explanation: The final route is shown.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
```

### Constraints

* n == grid.length
* n == grid\[i].length
* 1 \<= n \<= 50
* 0 \<= grid\[i]\[j] \< n\<sup>2\</sup>
* Each value grid\[i]\[j] is **unique**.

## Solution

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

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


class Solution:
    # Time: O(n^2 log n)
    # Space: O(n^2)
    def swim_in_water(self, grid: list[list[int]]) -> int:
        n = len(grid)
        # Minimize the maximum elevation encountered along the path.
        min_time: list[list[int | float]] = [[float("inf")] * n for _ in range(n)]
        min_time[0][0] = grid[0][0]
        # (cost, row, col) where cost = max elevation on path so far
        min_heap: list[tuple[int, int, int]] = [(grid[0][0], 0, 0)]

        while min_heap:
            time, row, col = heapq.heappop(min_heap)
            if row == n - 1 and col == n - 1:
                return time
            if time > min_time[row][col]:
                continue
            for delta_row, delta_col in ((0, 1), (0, -1), (1, 0), (-1, 0)):
                next_row, next_col = row + delta_row, col + delta_col
                if 0 <= next_row < n and 0 <= next_col < n:
                    arrival = max(time, grid[next_row][next_col])
                    if arrival < min_time[next_row][next_col]:
                        min_time[next_row][next_col] = arrival
                        heapq.heappush(min_heap, (arrival, next_row, next_col))
        return -1
```

## Complexity

| Time         | Space  |
| ------------ | ------ |
| O(n^2 log n) | O(n^2) |

## Tags

[NeetCode 150](/catalog/neetcode-150), [NeetCode 250](/catalog/neetcode-250), [NeetCode All](/catalog/neetcode).