1631. Path With Minimum Effort

# 1631. Path With Minimum Effort#

## 题目 #

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 updownleft, 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] <= 10^6`

## 解题思路 #

• 此题和第 778 题解题思路完全一致。在第 778 题中求的是最短连通时间。此题求的是连通路径下的最小体力值。都是求的最小值，只是 2 个值的意义不同罢了。
• 按照第 778 题的思路，本题也有多种解法。第一种解法是 DFS + 二分。先将题目变换一个等价问法。题目要求找到最小体力消耗值，也相当于问是否存在一个体力消耗值 x，只要大于等于 x，一定能连通。利用二分搜索来找到这个临界值。体力消耗值是有序的，此处满足二分搜索的条件。题目给定柱子高度是 [1,10^6]，所以体力值一定在 [0,10^6-1] 这个区间内。判断是否取中值的条件是用 DFS 或者 BFS 搜索 (0,0) 点和 (N-1, N-1) 点之间是否连通。时间复杂度：O(mnlogC)，其中 m 和 n 分别是地图的行数和列数，C 是格子的最大高度。C 最大为 10^6，所以 logC 常数也很小。空间复杂度 O(mn)。
• 第二种解法是并查集。将图中所有边按照权值从小到大进行排序，并依次加入并查集中。直到加入一条权值为 x 的边以后，左上角到右下角连通了。最小体力消耗值也就找到了。注意加入边的时候，只加入 `i-1``i``j-1``j` 这 2 类相邻的边。因为最小体力消耗意味着不走回头路。上下左右四个方向到达一个节点，只可能从上边和左边走过来。从下边和右边走过来肯定是浪费体力了。时间复杂度：O(mnlog(mn))，其中 m 和 n 分别是地图的行数和列数，图中的边数为 O(mn)。空间复杂度 O(mn)，即为存储所有边以及并查集需要的空间。

## 代码 #

``````package leetcode

import (
"sort"

"github.com/halfrost/leetcode-go/template"
)

var dir = [4][2]int{
{0, 1},
{1, 0},
{0, -1},
{-1, 0},
}

// 解法一 DFS + 二分
func minimumEffortPath(heights [][]int) int {
n, m := len(heights), len(heights[0])
visited := make([][]bool, n)
for i := range visited {
visited[i] = make([]bool, m)
}
low, high := 0, 1000000
for low < high {
threshold := low + (high-low)>>1
if !hasPath(heights, visited, 0, 0, threshold) {
low = threshold + 1
} else {
high = threshold
}
for i := range visited {
for j := range visited[i] {
visited[i][j] = false
}
}
}
return low
}

func hasPath(heights [][]int, visited [][]bool, i, j, threshold int) bool {
n, m := len(heights), len(heights[0])
if i == n-1 && j == m-1 {
return true
}
visited[i][j] = true
res := false
for _, d := range dir {
ni, nj := i+d[0], j+d[1]
if ni < 0 || ni >= n || nj < 0 || nj >= m || visited[ni][nj] || res {
continue
}
diff := abs(heights[i][j] - heights[ni][nj])
if diff <= threshold && hasPath(heights, visited, ni, nj, threshold) {
res = true
}
}
return res
}

func abs(a int) int {
if a < 0 {
a = -a
}
return a
}

func min(a, b int) int {
if a < b {
return a
}
return b
}

func max(a, b int) int {
if a < b {
return b
}
return a
}

// 解法二 并查集
func minimumEffortPath1(heights [][]int) int {
n, m, edges, uf := len(heights), len(heights[0]), []edge{}, template.UnionFind{}
uf.Init(n * m)
for i, row := range heights {
for j, h := range row {
id := i*m + j
if i > 0 {
edges = append(edges, edge{id - m, id, abs(h - heights[i-1][j])})
}
if j > 0 {
edges = append(edges, edge{id - 1, id, abs(h - heights[i][j-1])})
}
}
}
sort.Slice(edges, func(i, j int) bool { return edges[i].diff < edges[j].diff })
for _, e := range edges {
uf.Union(e.v, e.w)
if uf.Find(0) == uf.Find(n*m-1) {
return e.diff
}
}
return 0
}

type edge struct {
v, w, diff int
}
``````

Apr 8, 2023