# ---
# title: 1642. Furthest Building You Can Reach
# id: problem1642
# author: AquaIndigo
# date: 2020-11-26
# difficulty: Medium
# categories: Binary Search, Heap
# link: <https://leetcode.com/problems/furthest-building-you-can-reach/description/>
# hidden: true
# ---
# 
# 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:**
# 
# ![](https://assets.leetcode.com/uploads/2020/10/27/q4.gif)
# 
#     
#     
#     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
using DataStructures

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
