<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>

class Solution:
    def swimInWater(self, grid: List[List[int]]) -> int:
        n = len(grid)
        m = len(grid[0])
        in_range = lambda x, y: 0 <= x < n and 0 <= y < n
        directions = [[-1, 0], [0, -1], [1, 0], [0, 1]]
        visited = set()
        min_time = defaultdict(lambda x: float('inf'))
        queue = []
        
        heapq.heappush(queue, (grid[0][0], 0, 0))
        
        while queue:
            cost, cur_dx, cur_dy = heapq.heappop(queue)
            if (cur_dx, cur_dy) in visited:
                continue
            min_time[(cur_dx, cur_dy)] = cost
            visited.add((cur_dx, cur_dy))
            for dx, dy in directions:
                new_dx, new_dy = cur_dx + dx, cur_dy + dy
                if in_range(new_dx, new_dy):
                    heappush(queue, (max(cost, grid[new_dx][new_dy]), new_dx, new_dy))
        
        return min_time[(n - 1, m - 1)]
17500