1631-path-with-minimum-effort

Try it on leetcode

Description

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

Solution(Python)

class Solution:
    def minimumEffortPath(self, heights: List[List[int]]) -> int:
        return self.binarySearch(heights)

    # Time Complexity: O(ElogV) = O(M*N*log(M*N))
    # Space Complexity: O(M*N)
    def dijisktra(self, heights: List[List[int]]) -> int:
        m, n = len(heights), len(heights[0])
        dist = [[math.inf] * n for _ in range(m)]
        dist[0][0] = 0

        PQ = [(0, 0, 0)]
        DIRS = [(0, 1), (1, 0), (0, -1), (-1, 0)]

        while PQ:
            w, r, c = heappop(PQ)

            if w > dist[r][c]:
                continue
            if r == m - 1 and c == n - 1:
                return w

            for dx, dy in DIRS:
                nr, nc = r + dx, c + dy
                if self.isvalid(nr, nc, m, n):
                    newDist = max(w, abs(heights[nr][nc] - heights[r][c]))
                    if dist[nr][nc] > newDist:
                        dist[nr][nc] = newDist
                        heappush(PQ, (dist[nr][nc], nr, nc))

    def isvalid(self, x, y, m, n):
        return 0 <= x < m and 0 <= y < n

    # Time Complexity: O(M*N log(MaxHEight))
    # Space Complexity: O(m*n)
    def binarySearch(self, heights: List[List[int]]) -> int:
        m, n = len(heights), len(heights[0])
        DIR = [0, 1, 0, -1, 0]

        def dfs(r, c, visited, threadshold):
            if r == m - 1 and c == n - 1:
                return True  # Reach destination
            visited[r][c] = True
            for i in range(4):
                nr, nc = r + DIR[i], c + DIR[i + 1]
                if nr < 0 or nr == m or nc < 0 or nc == n or visited[nr][nc]:
                    continue
                if abs(heights[nr][nc] - heights[r][c]) <= threadshold and dfs(
                    nr, nc, visited, threadshold
                ):
                    return True
            return False

        def canReachDestination(threadshold):
            visited = [[False] * n for _ in range(m)]
            return dfs(0, 0, visited, threadshold)

        left = 0
        ans = right = 10**6
        while left <= right:
            mid = left + (right - left) // 2
            if canReachDestination(mid):
                right = mid - 1  # Try to find better result on the left side
                ans = mid
            else:
                left = mid + 1
        return ans