<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>

from heapq import heappush, heappop
class Solution:
    def minimumEffortPath(self, heights: List[List[int]]) -> int:
        m, n = len(heights), len(heights[0])
        effort = [[float('inf')] * n for _ in range(m)]
        effort[0][0] = 0
        pq = []
        heappush(pq, State(0, 0, 0))
        while pq:
            cur_state = heappop(pq)
            cur_x = cur_state.x 
            cur_y = cur_state.y 
            cur_effort = cur_state.effort
            if cur_x == m - 1 and cur_y == n - 1:
                return cur_effort
            if effort[cur_x][cur_y] < cur_effort:
                continue
            effort[cur_x][cur_y] = cur_effort
            for neighbor_x, neighbor_y in self.get_neighbors(heights, cur_x, cur_y):
                 next_effort = max(effort[cur_x][cur_y], abs(heights[cur_x][cur_y] - heights[neighbor_x][neighbor_y]))
                 if effort[neighbor_x][neighbor_y] > next_effort:
                    effort[neighbor_x][neighbor_y] = next_effort
                    heappush(pq, State(neighbor_x, neighbor_y, next_effort))
        
        return -1

    
    def get_neighbors(self, graph: List[List[int]], i: int, j: int) -> List[int]:
        m, n = len(graph), len(graph[0])
        dirs = [(-1, 0), (1, 0), (0, -1), (0, 1)]
        neighbors = []
        for di, dj in dirs:
            next_i, next_j = i + di, j + dj
            if next_i < 0 or next_i >= m or next_j < 0 or next_j >= n:
                continue
            neighbors.append((next_i, next_j))
        return neighbors

    


class State:
    def __init__(self, x: int, y: int, effort: int):
        self.x = x
        self.y = y
        self.effort = effort
    

    def __lt__(self, other) -> bool:
        return self.effort < other.effort
        
18100