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 differencein 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: “Lazy BFS / DP”

dp[y][x] = min(max(dp[ty][tx], abs(h[ty][tx] – h[y][x]))) (x, y) and (tx, ty) are neighbors
repeat this process for at most rows * cols times.
if dp does not change after one round which means we found the optimal solution and can break earlier.

Time complexity: O(n^2*m^2))
Space complexity: O(nm)

## Solution 2: Binary Search + BFS

Use binary search to guess a cost and then check whether there is path that is under the cost.

Time complexity: O(mn*log(max(h) – min(h)))
Space complexity: O(mn)

## Solution 3: Dijkstra

Time complexity: O(mnlog(mn))
Space complexity: O(mn)

## C++

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