<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:
        rows, cols = len(heights), len(heights[0])
        
        # Helper function to find the root of a set in Union-Find
        def find(parent, x):
            if parent[x] != x:
                parent[x] = find(parent, parent[x])
            return parent[x]
        
        # Helper function to union two sets in Union-Find
        def union(parent, rank, x, y):
            rootX = find(parent, x)
            rootY = find(parent, y)
            if rootX != rootY:
                if rank[rootX] > rank[rootY]:
                    parent[rootY] = rootX
                elif rank[rootX] < rank[rootY]:
                    parent[rootX] = rootY
                else:
                    parent[rootY] = rootX
                    rank[rootX] += 1
        
        # Edge list of all possible connections with their respective efforts
        edges = []
        for r in range(rows):
            for c in range(cols):
                if r > 0:  # Connect with the cell above
                    edges.append((abs(heights[r][c] - heights[r-1][c]), r * cols + c, (r-1) * cols + c))
                if c > 0:  # Connect with the cell to the left
                    edges.append((abs(heights[r][c] - heights[r][c-1]), r * cols + c, r * cols + (c-1)))
        
        # Sort edges by effort in ascending order
        edges.sort()
        
        # Union-Find setup
        parent = [i for i in range(rows * cols)]
        rank = [0] * (rows * cols)
        
        # Iterate over sorted edges, union sets and check if start and end are connected
        for effort, x, y in edges:
            union(parent, rank, x, y)
            # If top-left and bottom-right are connected, return the current effort
            if find(parent, 0) == find(parent, rows * cols - 1):
                return effort
        
        return 0

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