LeetCode 1631. Path With Minimum Effort

Description

https://leetcode.com/problems/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 updownleft, 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

Python Solution

Use binary search to search for minimum effort, within each binary search, use depth-first search to find if can find a path within the effort limit.

class Solution:
    def minimumEffortPath(self, heights: List[List[int]]) -> int:        
        start = 0
        end = 10000000
        
        while start + 1 < end:
            mid = (start + end) // 2

            if self.dfs(heights, set(), 0, 0, mid):
                end = mid
            else:
                start = mid
                
        if self.dfs(heights, set(), 0, 0, start):
            return start
     
        if self.dfs(heights, set(), 0, 0, end):        
            return end
        
        return -1
    
    
    def dfs(self, heights, visited, x, y, limit):
        
        if x == len(heights) - 1 and y == len(heights[0]) - 1:
            return True
        
        visited.add((x, y))

        DIRECTIONS = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        
        for dx, dy in DIRECTIONS:
            next_x = x + dx
            next_y = y + dy
            
            
            if not self.is_valid(heights, next_x, next_y):
                continue
            
            if (next_x, next_y) in visited:
                continue
            
            if abs(heights[next_x][next_y] - heights[x][y]) > limit:
                continue
            
            if self.dfs(heights, visited, next_x, next_y, limit):
                return True
            
        return False
        
    def is_valid(self, heights, x, y):
        if not (0 <= x < len(heights) and 0 <= y < len(heights[0])):
            return False
        
        return True
            
            
    
  • Time Complexity: O(MN), where M is the number of rows of the matrix, N is the number of columns of the matrix.
  • Space Complexity: O(MN).

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