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].

class Solution {
public int swimInWater(int[][] grid) {
}
}