1631. Path With Minimum Effort

Description

You are a hiker preparing for an upcoming hike. You are given heights, a 2D array of size rows x columns, where heights[row][col] represents the height of cell (row, col). You are situated in the top-left cell, (0, 0), and you hope to travel to the bottom-right cell, (rows-1, columns-1) (i.e., 0-indexed). You can move up, down, left, or right, and you wish to find a route that requires the minimum effort.

A route's effort is the maximum absolute difference in heights between two consecutive cells of the route.

Return the minimum effort required to travel from the top-left cell to the bottom-right cell.

 

Example 1:

Input: heights = [[1,2,2],[3,8,2],[5,3,5]]
Output: 2
Explanation: 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.

Example 2:

Input: heights = [[1,2,3],[3,8,4],[5,3,5]]
Output: 1
Explanation: 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].

Example 3:

Input: 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]]
Output: 0
Explanation: This route does not require any effort.

 

Constraints:

• rows == heights.length
• columns == heights[i].length
• 1 <= rows, columns <= 100
• 1 <= heights[i][j] <= 106

Solutions

Solution 1: Union-Find

For this problem, we can treat each cell as a node in a graph, and the absolute difference in height between two adjacent cells as the weight of the edge. Therefore, this problem is to solve the connectivity problem from the top-left node to the bottom-right node.

We first construct a set of edges, then sort them in ascending order of edge weight, and add edges one by one until the top-left node and the bottom-right node are connected. At this point, the weight of the edge is the minimum physical consumption value required by the problem.

The time complexity is $O(m \times n \times \log(m \times n))$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns in the two-dimensional array, respectively.

Python3JavaC++GoTypeScript

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

class UnionFind {
    private final int[] p;
    private final int[] size;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
    }

    public int find(int x) {
        if (p[x] != x) {
            p[x]