Path With Minimum Effort
Tags: leetcode, dsa, theabbie
Published: 2022-06-07 09:30:42

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`

**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