<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 UnionFind:
    def __init__(self, size):
        self.parent = [i for i in range(size)]
        self.rank = [0 for _ in range(size)]
    
    def find(self, i):
        if self.parent[i] == i: return i
        self.parent[i] = self.find(self.parent[i])
        return self.parent[i]
    
    def set_union(self, i, j):
        i = self.find(i)
        j = self.find(j)

        if i == j: return False
        if self.rank[i] < self.rank[j]:
            self.parent[i] = j
        elif self.rank[i] > self.rank[j]:
            self.parent[j] = i
        else:
            self.parent[j] = i
            self.rank[i] += 1
        return True

class Solution:
    def minimumEffortPath(self, heights: List[List[int]]) -> int:
        edges = []
        m, n = len(heights), len(heights[0])
        for i in range(m):
            for j in range(n):
                if i > 0:
                    diff = abs(heights[i][j] - heights[i-1][j])
                    edges.append((diff, i*n + j, (i-1)*n + j))
                if j > 0:
                    diff = abs(heights[i][j] - heights[i][j-1])
                    edges.append((diff, i*n + j, i*n + j-1))

        edges.sort()
        union_find = UnionFind(size=m*n)

        for diff, x, y in edges:
            union_find.set_union(x, y)
            if union_find.find(0) == union_find.find(n*m - 1):
                return diff
        return 0
20400