g0701_0800.s0778_swim_in_rising_water.readme.md Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java17 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.24

778\. Swim in Rising Water

Hard

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

The rain starts to fall. At time `t`, the depth of the water everywhere is `t`. 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 least time until you can reach the bottom right square_ `(n - 1, n - 1)` _if you start at the top left square_ `(0, 0)`.

**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] < n²
*   Each value `grid[i][j]` is **unique**.