## DEV Community

Abhishek Chaudhary

Posted on

# 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: 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`

SOLUTION:

``````class Solution:
def isPossible(self, heights, i, j, m, n, visited, k):
if (i, j) == (m - 1, n - 1):
return True
for (x, y) in [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]:
if 0 <= x < m and 0 <= y < n and (x, y) not in visited:
effort = abs(heights[x][y] - heights[i][j])
if effort <= k:
if self.isPossible(heights, x, y, m, n, visited, k):
return True
return False

def minimumEffortPath(self, heights: List[List[int]]) -> int:
m = len(heights)
n = len(heights[0])
beg = 0
end = 1
while not self.isPossible(heights, 0, 0, m, n, {(0, 0)}, end):
end = end << 1
while beg <= end:
mid = (beg + end) // 2
iscurrpossible = self.isPossible(heights, 0, 0, m, n, {(0, 0)}, mid)
isprevpossible = self.isPossible(heights, 0, 0, m, n, {(0, 0)}, mid - 1)
if not isprevpossible and iscurrpossible:
return mid
elif beg == end:
break
elif not isprevpossible and not iscurrpossible:
beg = mid + 1
elif isprevpossible and iscurrpossible:
end = mid
return beg
``````