# LeetCode 1631. Path With Minimum Effort

## 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 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] <= 106`

## 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

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).