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

<p>The rain starts to fall. At time <code>t</code>, the depth of the water everywhere is <code>t</code>. 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 <code>t</code>. You can swim infinite distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.</p>

<p>Return <em>the least time until you can reach the bottom right square </em><code>(n - 1, n - 1)</code><em> if you start at the top left square </em><code>(0, 0)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim1-grid.jpg" style="width: 164px; height: 165px;" />
<pre>
<strong>Input:</strong> grid = [[0,2],[1,3]]
<strong>Output:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim2-grid-1.jpg" style="width: 404px; height: 405px;" />
<pre>
<strong>Input:</strong> 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]]
<strong>Output:</strong> 16
<strong>Explanation:</strong> The final route is shown.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;&nbsp;n<sup>2</sup></code></li>
	<li>Each value <code>grid[i][j]</code> is <strong>unique</strong>.</li>
</ul>

# Question similar to "Path with Minimum Effort" 
class Solution:
    def swimInWater(self, grid: List[List[int]]) -> int:

        # Getting rows and cols
        r=len(grid)
        c=len(grid[0])

        # 2D array to store minimum elevation to reach there
        cache=[[math.inf for _ in range(c)] for _ in range(r)]

        # Directions to traverse
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

        # Minheap storing (elevation,x co-ordinat, y co-ordinate)
        minheap=[(grid[0][0],0,0)]

        # Remains same, as this is start postion
        cache[0][0] = grid[0][0]

        while minheap:
            # Popping position with miniumm elevation
            elevation,i,j=heapq.heappop(minheap)

            # Base case to reach end
            if i==(r-1) and j==(c-1):
                return elevation
            
            # Traversing 4 directions
            for dx, dy in directions:
                nx, ny = i + dx, j + dy
                
                if 0 <= nx < r and 0 <= ny < c:

                    # "new_elevation" keeping track of max encountered till now
                    new_elevation=max(grid[nx][ny],elevation)

                    # Update it in cache as well as add in heap if "new_elevation" is less
                    # than one in cache
                    if new_elevation < cache[nx][ny]:
                        cache[nx][ny] = new_elevation
                        heappush(minheap, (new_elevation, nx, ny))

            
        
16700