1642. Furthest Building You Can Reach

You are given an integer array heights representing the heights of buildings, some bricks, and some ladders.

You start your journey from building 0 and move to the next building by possibly using bricks or ladders.

While moving from building i to building i+1 ( 0-indexed ),

• If the current building's height is greater than or equal to the next building's height, you do not need a ladder or bricks.
• If the current building's height is less than the next building's height, you can either use one ladder or (h[i+1] - h[i]) bricks.

Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.

Example 1:

Input: heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1
Output: 4
Explanation: Starting at building 0, you can follow these steps:
- Go to building 1 without using ladders nor bricks since 4 >= 2.
- Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 < 7.
- Go to building 3 without using ladders nor bricks since 7 >= 6.
- Go to building 4 using your only ladder. You must use either bricks or ladders because 6 < 9.
It is impossible to go beyond building 4 because you do not have any more bricks or ladders.

Example 2:

Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
Output: 7

Example 3:

Input: heights = [14,3,19,3], bricks = 17, ladders = 0
Output: 3

Constraints:

• 1 <= heights.length <= 105
• 1 <= heights[i] <= 106
• 0 <= bricks <= 109
• 0 <= ladders <= heights.length

# @lc code=start
using LeetCode

function furthest_building(heights::Vector{Int}, bricks::Int, ladders::Int)
    q = Int[]
    for i in 1:(length(heights) - 1)
        diff = heights[i + 1] - heights[i]
        if diff > 0
            if diff > bricks && ladders == 0
                return i - 1
            end
            if bricks >= diff
                heappush!(q, diff)
                bricks -= diff
            else
                if !isempty(q) && q[1] > diff
                    bricks += heappop!(q) - diff
                    heappush!(q, diff)
                end
                ladders -= 1
            end
        end
    end
    return length(heights) - 1
end
# @lc code=end

furthest_building (generic function with 1 method)

This page was generated using DemoCards.jl and Literate.jl.