1631. Path With Minimum Effort

Medium

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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

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