Path With Minimum Effort
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1600-1699/1631.Path%20With%20Minimum%20Effort/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1600-1699/1631.Path%20With%20Minimum%20Effort/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1600-1699/1631.Path%20With%20Minimum%20Effort/images/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>
Depth-First Search; Breadth-First Search; Union Find; Array; Binary Search; Matrix; Heap (Priority Queue)
class UnionFind:
    def __init__(self, n):
        self.p = list(range(n))
        self.size = [1] * n

    def find(self, x):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]

    def union(self, a, b):
        pa, pb = self.find(a), self.find(b)
        if pa == pb:
            return False
        if self.size[pa] > self.size[pb]:
            self.p[pb] = pa
            self.size[pa] += self.size[pb]
        else:
            self.p[pa] = pb
            self.size[pb] += self.size[pa]
        return True

    def connected(self, a, b):
        return self.find(a) == self.find(b)


class Solution:
    def minimumEffortPath(self, heights: List[List[int]]) -> int:
        m, n = len(heights), len(heights[0])
        uf = UnionFind(m * n)
        e = []
        dirs = (0, 1, 0)
        for i in range(m):
            for j in range(n):
                for a, b in pairwise(dirs):
                    x, y = i + a, j + b
                    if 0 <= x < m and 0 <= y < n:
                        e.append(
                            (abs(heights[i][j] - heights[x][y]), i * n + j, x * n + y)
                        )
        e.sort()
        for h, a, b in e:
            uf.union(a, b)
            if uf.connected(0, m * n - 1):
                return h
        return 0
