<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:
        ROW = len(heights)
        COL = len(heights[0])
        DIRS = [(0, 1), (0, -1), (-1, 0), (1, 0)]
        visted = set()
        differences = [[inf] * COL for _ in range(ROW)]

        priorityQueue = [(0, (0, 0))]
        differences[0][0] = 0

        while priorityQueue and len(visted) < ROW * COL:
            currDiff, currNode = heapq.heappop(priorityQueue)

            # if currNode == (ROW - 1, COL - 1):
            #     return differences[ROW - 1][COL - 1]

            if currNode not in visted:
                visted.add(currNode)
                
                for dR, dC in DIRS:
                    nR = currNode[0] + dR
                    nC = currNode[1] + dC
                    if 0 <= nR < ROW and 0 <= nC < COL and (nR, nC) not in visted:
                        newDiff = abs(heights[nR][nC] - heights[currNode[0]][currNode[1]])
                        maxDiff = max(newDiff, differences[currNode[0]][currNode[1]])
                        differences[nR][nC] = min( differences[nR][nC], maxDiff )
                        heappush( priorityQueue, (maxDiff, (nR, nC)))

        return differences[-1][-1]
19500