<p>You are a hiker preparing for an upcoming hike. You are given <code>heights</code>, a 2D array of size <code>rows x columns</code>, where <code>heights[row][col]</code> represents the height of cell <code>(row, col)</code>. You are situated in the top-left cell, <code>(0, 0)</code>, and you hope to travel to the bottom-right cell, <code>(rows-1, columns-1)</code> (i.e.,&nbsp;<strong>0-indexed</strong>). You can move <strong>up</strong>, <strong>down</strong>, <strong>left</strong>, or <strong>right</strong>, and you wish to find a route that requires the minimum <strong>effort</strong>.</p>

<p>A route&#39;s <strong>effort</strong> is the <strong>maximum absolute difference</strong><strong> </strong>in heights between two consecutive cells of the route.</p>

<p>Return <em>the minimum <strong>effort</strong> required to travel from the top-left cell to the bottom-right cell.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/10/04/ex1.png" style="width: 300px; height: 300px;" /></p>

<pre>
<strong>Input:</strong> heights = [[1,2,2],[3,8,2],[5,3,5]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/10/04/ex2.png" style="width: 300px; height: 300px;" /></p>

<pre>
<strong>Input:</strong> heights = [[1,2,3],[3,8,4],[5,3,5]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/04/ex3.png" style="width: 300px; height: 300px;" />
<pre>
<strong>Input:</strong> heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> This route does not require any effort.
</pre>

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

<ul>
	<li><code>rows == heights.length</code></li>
	<li><code>columns == heights[i].length</code></li>
	<li><code>1 &lt;= rows, columns &lt;= 100</code></li>
	<li><code>1 &lt;= heights[i][j] &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
    def minimumEffortPath(self, heights: List[List[int]]) -> int:
        # djikstras
        # instead of keeping the sum, we just keep the maximum abs diff between cells
        # and ofc the visited arr and the queue

        visited = set() # (i, j)
        min_vals = defaultdict(int) # (i, j): dist
        q = [] # (max_abs_diff, (i, j))
        q.append((0, (0, 0)))
        heapq.heapify(q)

        while q:
            curr_d, curr_ind = heapq.heappop(q)
            i, j = curr_ind

            if i == len(heights) - 1 and j == len(heights[0]) - 1:
                return curr_d
 
            visited.add(curr_ind)
            # min_vals[curr_ind] = curr_d

            for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                new_i, new_j = i + dx, j + dy
                if new_i < 0 or new_i >= len(heights) or new_j < 0 or new_j >= len(heights[new_i]) or (new_i, new_j) in visited:
                    continue
                curr_diff = max(abs(heights[new_i][new_j] - heights[i][j]), curr_d)
                if (new_i, new_j) not in min_vals or min_vals[(new_i, new_j)] > curr_diff:
                    min_vals[(new_i, new_j)] = curr_diff
                    heapq.heappush(q, (curr_diff, (new_i, new_j)))
        
        return -1

# class Solution:
#     def minimumEffortPath(self, heights: List[List[int]]) -> int:
#         row = len(heights)
#         col = len(heights[0])
#         difference_matrix = [[math.inf]*col for _ in range(row)]
#         difference_matrix[0][0] = 0
#         visited = [[False]*col for _ in range(row)]
#         queue = [(0, 0, 0)]  # difference, x, y
#         while queue:
#             difference, x, y = heapq.heappop(queue)
#             visited[x][y] = True
#             for dx, dy in [[0, 1], [1, 0], [0, -1], [-1, 0]]:
#                 adjacent_x = x + dx
#                 adjacent_y = y + dy
#                 if 0 <= adjacent_x < row and 0 <= adjacent_y < col and not visited[adjacent_x][adjacent_y]:
#                     current_difference = abs(heights[adjacent_x][adjacent_y]-heights[x][y])
#                     max_difference = max(current_difference, difference_matrix[x][y])
#                     if difference_matrix[adjacent_x][adjacent_y] > max_difference:
#                         difference_matrix[adjacent_x][adjacent_y] = max_difference
#                         heapq.heappush(queue, (max_difference, adjacent_x, adjacent_y))
#         return difference_matrix[-1][-1]

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