## Description

https://leetcode.com/problems/path-with-minimum-effort/

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:2Explanation: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:1Explanation: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:0Explanation: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}

## Python Solution

Use binary search to search for minimum effort, within each binary search, use depth-first search to find if can find a path within the effort limit.

```
class Solution:
def minimumEffortPath(self, heights: List[List[int]]) -> int:
start = 0
end = 10000000
while start + 1 < end:
mid = (start + end) // 2
if self.dfs(heights, set(), 0, 0, mid):
end = mid
else:
start = mid
if self.dfs(heights, set(), 0, 0, start):
return start
if self.dfs(heights, set(), 0, 0, end):
return end
return -1
def dfs(self, heights, visited, x, y, limit):
if x == len(heights) - 1 and y == len(heights[0]) - 1:
return True
visited.add((x, y))
DIRECTIONS = [(0, 1), (1, 0), (0, -1), (-1, 0)]
for dx, dy in DIRECTIONS:
next_x = x + dx
next_y = y + dy
if not self.is_valid(heights, next_x, next_y):
continue
if (next_x, next_y) in visited:
continue
if abs(heights[next_x][next_y] - heights[x][y]) > limit:
continue
if self.dfs(heights, visited, next_x, next_y, limit):
return True
return False
def is_valid(self, heights, x, y):
if not (0 <= x < len(heights) and 0 <= y < len(heights[0])):
return False
return True
```

- Time Complexity: O(MN), where M is the number of rows of the matrix, N is the number of columns of the matrix.
- Space Complexity: O(MN).