## 778. Swim in Rising Water

On an N x N `grid`

, each square `grid[i][j]`

represents the elevation at that point `(i,j)`

.

Now 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 distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.

You start at the top left square `(0, 0)`

. What is the least time until you can reach the bottom right square `(N-1, N-1)`

?

**Example 1:**

Input:[[0,2],[1,3]]Output:3Explanation: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:**

Input:[[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:16Explanation:0 1 2 3 424 23 22 21512 13 14 15 161117 18 19 2010 9 8 7 6The final route is marked in bold. We need to wait until time 16 so that (0, 0) and (4, 4) are connected.

**Note:**

`2 <= N <= 50`

.- grid[i][j] is a permutation of [0, ..., N*N - 1].

## Rust Solution

```
struct Solution;
use std::cmp::Reverse;
use std::collections::BinaryHeap;
struct UnionFind {
parent: Vec<usize>,
n: usize,
}
impl UnionFind {
fn new(n: usize) -> Self {
let parent = (0..n).collect();
UnionFind { parent, n }
}
fn find(&mut self, i: usize) -> usize {
let j = self.parent[i];
if i == j {
i
} else {
let k = self.find(j);
self.parent[i] = k;
k
}
}
fn union(&mut self, i: usize, j: usize) {
let i = self.find(i);
let j = self.find(j);
if i != j {
self.parent[i] = j;
}
}
}
impl Solution {
fn swim_in_water(grid: Vec<Vec<i32>>) -> i32 {
let n = grid.len();
let mut uf = UnionFind::new(n * n);
let mut queue: BinaryHeap<(Reverse<i32>, usize, usize)> = BinaryHeap::new();
for i in 0..n {
for j in 0..n {
queue.push((Reverse(grid[i][j]), i, j));
}
}
while let Some((Reverse(t), r, c)) = queue.pop() {
let i = r * n + c;
if grid[r][c] <= t && r > 0 && grid[r - 1][c] <= t {
let j = (r - 1) * n + c;
uf.union(i, j);
}
if grid[r][c] <= t && r + 1 < n && grid[r + 1][c] <= t {
let j = (r + 1) * n + c;
uf.union(i, j);
}
if grid[r][c] <= t && c > 0 && grid[r][c - 1] <= t {
let j = r * n + (c - 1);
uf.union(i, j);
}
if grid[r][c] <= t && c + 1 < n && grid[r][c + 1] <= t {
let j = r * n + (c + 1);
uf.union(i, j);
}
if uf.find(0) == uf.find(n * n - 1) {
return t;
}
}
0
}
}
#[test]
fn test() {
let grid = vec_vec_i32![[0, 2], [1, 3]];
let res = 3;
assert_eq!(Solution::swim_in_water(grid), res);
let grid = vec_vec_i32![
[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]
];
let res = 16;
assert_eq!(Solution::swim_in_water(grid), res);
}
```

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