<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)
        directions = [(0,1),(0,-1),(1,0),(-1,0)]
        @cache
        def inbounds(pt):
            return N > pt[0] >= 0 <= pt[1] < N

        minHeap = [(grid[0][0], (0,0))]  #(time, (r,c))
        visited = set()

        while minHeap:

            t,pt = heapq.heappop(minHeap)
            if pt == (N-1,N-1):
                return t

            if pt in visited:
                continue
            visited.add(pt)

            neighbors = [(pt[0]+d[0],pt[1]+d[1]) for d in directions]
            for n in neighbors:
                if inbounds(n):
                    a,b = n
                    heapq.heappush(minHeap, (max(t, grid[a][b]), n))

     

    def swimInWaterKD(self, grid: List[List[int]]) -> int:
        
        n=len(grid)
        directions = [(0,1),(0,-1),(1,0),(-1,0)]

        minHeap = [(grid[0][0], (0,0))]
        shortest = {}

        # (r=row,c=column),d=depth
        while minHeap:

            (d, (r,c)) = heapq.heappop(minHeap)

            if r==n-1 and c==n-1:
                #print(shortest)
                return d

            if (r,c) in shortest:
                continue
            shortest[(r,c)] = d

            
            for neigh in [(r+a[0],c+a[1]) for a in directions]:
                if n > neigh[0] >= 0 <= neigh[1] < n:

                    if not neigh in shortest:
                        #pt = (neigh, d + grid[neigh[0]][neigh[1]])
                        pt = (max(d, grid[neigh[0]][neigh[1]]), neigh)
                        heapq.heappush(minHeap, pt)

        return -1
17800