cpp-flashcards/org/study_deck_02/dsa/advanced-graphs/0778-swim-in-rising-water.org

#+PROPERTY: STUDY_DECK_02
* TODO 0778. Swim In Rising Water :hard:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*0778. Swim In Rising Water][0778. Swim In Rising Water]]
:END:

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)~.

*Example 1:*


#+begin_src
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.
#+end_src


*Example 2:*


#+begin_src
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.
#+end_src


*Constraints:*

 - ~n == grid.length~

 - ~n == grid[i].length~

 - ~1 <= n <= 50~

 - ~0 <= grid[i][j] < n^{2}~

 - Each value ~grid[i][j]~ is *unique*.

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python
class Solution:
    def swimInWater(self, grid: List[List[int]]) -> int:
#+end_src

** TODO C++
#+begin_src cpp
class Solution {
public:
    int swimInWater(vector<vector<int>>& grid) {
        
    }
};
#+end_src