# 1631. Path With Minimum Effort

#### Medium

***

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:**

![](https://assets.leetcode.com/uploads/2020/10/04/ex1.png)

```
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:**

![](https://assets.leetcode.com/uploads/2020/10/04/ex2.png)

```
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:**

![](https://assets.leetcode.com/uploads/2020/10/04/ex3.png)

```
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
class Solution:
    def minimumEffortPath(self, heights: List[List[int]]) -> int:
        #.
        self.rows = len(heights)
        self.cols = len(heights[0])
        self.grid = heights
        low = 0
        high = 10**6
        while low < high:
            mid = low + (high-low) // 2
            if self.bfs(mid):
                high = mid
            else:
                low = mid + 1
        return low
        
        
    def bfs(self, threshold):
        visited = {(0,0)}
        dq = deque([(0,0)])
        while dq:
            x, y = dq.popleft()
            # Reached bottom right
            if (x,y) == (self.rows-1, self.cols-1):
                return True
            for r,c in [(x+1,y), (x,y+1), (x-1,y), (x,y-1)]:
                if (r,c) not in visited and (0 <= r < self.rows and 0 <= c < self.cols) and abs(self.grid[r][c] - self.grid[x][y]) <= threshold:
                    visited.add((r,c))
                    dq.append((r,c))
        return False

```