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:
                    visited.add((x, y))
                    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