**Swim in Rising Water ** 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:**

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:**

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:16Explanation: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`

^{2}- Each value
`grid[i][j]`

is**unique**.

### Swim in Rising Water Solutions

✅**Time:** O(n)

✅**Space:** O(n)

**C**++

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**Python**

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