1631. Path With Minimum Effort

Source code notebook Author Update time

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
# @lc code=start
using LeetCode

function minimum_effort_path(heights::Vector{Vector{Int}})
    m, n = length(heights), length(heights[1])
    ds = IntDisjointSets(m * n)
    map = SortedMultiDict{Int, Pair{Int, Int}}()
    for i in 1:m-1
        for j in 1:n-1
            insert!(map, abs(heights[i][j] - heights[i+1][j]), Pair((i - 1) * n + j, i * n + j))
            insert!(map, abs(heights[i][j] - heights[i][j+1]), Pair((i - 1) * n + j, (i - 1) * n + j + 1))
        end
    end
    for i in 1:m-1
        insert!(map, abs(heights[i][n] - heights[i+1][n]), Pair(i * n, (i + 1) * n))
    end
    for j in 1:n-1
        insert!(map, abs(heights[m][j] - heights[m][j+1]), Pair((m-1) * n + j, (m-1) * n + j + 1))
    end
    for (dist, edge) in map
        union!(ds, edge.first, edge.second)
        if in_same_set(ds, 1, m * n)
            return dist
        end
    end
end
# @lc code=end
minimum_effort_path (generic function with 1 method)

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