# Posts tagged as “heap”

You are given a 2D integer array orders, where each orders[i] = [pricei, amounti, orderTypei] denotes that amountiorders have been placed of type orderTypei at the price pricei. The orderTypei is:

• 0 if it is a batch of buy orders, or
• 1 if it is a batch of sell orders.

Note that orders[i] represents a batch of amounti independent orders with the same price and order type. All orders represented by orders[i] will be placed before all orders represented by orders[i+1] for all valid i.

There is a backlog that consists of orders that have not been executed. The backlog is initially empty. When an order is placed, the following happens:

• If the order is a buy order, you look at the sell order with the smallest price in the backlog. If that sell order’s price is smaller than or equal to the current buy order’s price, they will match and be executed, and that sell order will be removed from the backlog. Else, the buy order is added to the backlog.
• Vice versa, if the order is a sell order, you look at the buy order with the largest price in the backlog. If that buy order’s price is larger than or equal to the current sell order’s price, they will match and be executed, and that buy order will be removed from the backlog. Else, the sell order is added to the backlog.

Return the total amount of orders in the backlog after placing all the orders from the input. Since this number can be large, return it modulo 109 + 7.

Example 1: Input: orders = [[10,5,0],[15,2,1],[25,1,1],[30,4,0]]
Output: 6
Explanation: Here is what happens with the orders:
- 5 orders of type buy with price 10 are placed. There are no sell orders, so the 5 orders are added to the backlog.
- 2 orders of type sell with price 15 are placed. There are no buy orders with prices larger than or equal to 15, so the 2 orders are added to the backlog.
- 1 order of type sell with price 25 is placed. There are no buy orders with prices larger than or equal to 25 in the backlog, so this order is added to the backlog.
- 4 orders of type buy with price 30 are placed. The first 2 orders are matched with the 2 sell orders of the least price, which is 15 and these 2 sell orders are removed from the backlog. The 3rd order is matched with the sell order of the least price, which is 25 and this sell order is removed from the backlog. Then, there are no more sell orders in the backlog, so the 4th order is added to the backlog.
Finally, the backlog has 5 buy orders with price 10, and 1 buy order with price 30. So the total number of orders in the backlog is 6.


Example 2: Input: orders = [[7,1000000000,1],[15,3,0],[5,999999995,0],[5,1,1]]
Output: 999999984
Explanation: Here is what happens with the orders:
- 109 orders of type sell with price 7 are placed. There are no buy orders, so the 109 orders are added to the backlog.
- 3 orders of type buy with price 15 are placed. They are matched with the 3 sell orders with the least price which is 7, and those 3 sell orders are removed from the backlog.
- 999999995 orders of type buy with price 5 are placed. The least price of a sell order is 7, so the 999999995 orders are added to the backlog.
- 1 order of type sell with price 5 is placed. It is matched with the buy order of the highest price, which is 5, and that buy order is removed from the backlog.
Finally, the backlog has (1000000000-3) sell orders with price 7, and (999999995-1) buy orders with price 5. So the total number of orders = 1999999991, which is equal to 999999984 % (109 + 7).


Constraints:

• 1 <= orders.length <= 105
• orders[i].length == 3
• 1 <= pricei, amounti <= 109
• orderTypei is either 0 or 1.

## Solution: Treemap / PriorityQueue / Heap

sell backlog: min heap
Trade happens between the tops of two queues.

Time complexity: O(nlogn)
Space complexity: O(n)

## C++

There is a school that has classes of students and each class will be having a final exam. You are given a 2D integer array classes, where classes[i] = [passi, totali]. You know beforehand that in the ith class, there are totali total students, but only passi number of students will pass the exam.

You are also given an integer extraStudents. There are another extraStudents brilliant students that are guaranteed to pass the exam of any class they are assigned to. You want to assign each of the extraStudents students to a class in a way that maximizes the average pass ratio across all the classes.

The pass ratio of a class is equal to the number of students of the class that will pass the exam divided by the total number of students of the class. The average pass ratio is the sum of pass ratios of all the classes divided by the number of the classes.

Return the maximum possible average pass ratio after assigning the extraStudents students. Answers within 10-5 of the actual answer will be accepted.

Example 1:

Input: classes = [[1,2],[3,5],[2,2]], extraStudents = 2
Output: 0.78333
Explanation: You can assign the two extra students to the first class. The average pass ratio will be equal to (3/4 + 3/5 + 2/2) / 3 = 0.78333.


Example 2:

Input: classes = [[2,4],[3,9],[4,5],[2,10]], extraStudents = 4
Output: 0.53485


Constraints:

• 1 <= classes.length <= 105
• classes[i].length == 2
• 1 <= passi <= totali <= 105
• 1 <= extraStudents <= 105

## Solution: Greedy + Heap

Sort by the ratio increase potential (p + 1) / (t + 1) – p / t.

Time complexity: O((m+n)logn)
Space complexity: O(n)

## Python3

There is a special kind of apple tree that grows apples every day for n days. On the ith day, the tree grows apples[i] apples that will rot after days[i] days, that is on day i + days[i] the apples will be rotten and cannot be eaten. On some days, the apple tree does not grow any apples, which are denoted by apples[i] == 0 and days[i] == 0.

You decided to eat at most one apple a day (to keep the doctors away). Note that you can keep eating after the first n days.

Given two integer arrays days and apples of length n, return the maximum number of apples you can eat.

Example 1:

Input: apples = [1,2,3,5,2], days = [3,2,1,4,2]
Output: 7
Explanation: You can eat 7 apples:
- On the first day, you eat an apple that grew on the first day.
- On the second day, you eat an apple that grew on the second day.
- On the third day, you eat an apple that grew on the second day. After this day, the apples that grew on the third day rot.
- On the fourth to the seventh days, you eat apples that grew on the fourth day.


Example 2:

Input: apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
Output: 5
Explanation: You can eat 5 apples:
- On the first to the third day you eat apples that grew on the first day.
- Do nothing on the fouth and fifth days.
- On the sixth and seventh days you eat apples that grew on the sixth day.


Constraints:

• apples.length == n
• days.length == n
• 1 <= n <= 2 * 104
• 0 <= apples[i], days[i] <= 2 * 104
• days[i] = 0 if and only if apples[i] = 0.

## Solution: PriorityQueue

Sort by rotten day in ascending order, only push onto the queue when that day has come (be able to grow apples).

Time complexity: O((n+ d)logn)
Space complexity: O(n)

## C++

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

## Solution 0: DFS

Time complexity: O(2^n)
Space complexity: O(n)

AC but should be TLE

## Solution 1: Binary Search + Greedy

Guess we can reach to m, sort the height differences from 0~m. Use ladders for larger values and use bricks for smallest values left.

Time complexity: O(nlogn)
Space complexity: O(n)

## Solution 2: Min heap

Use a min heap to store all the height differences ( > 0) so far, if heap size is greater than ladders, which means we have to use bricks, extract the smallest value and subtract the bricks.

Time complexity: O(nlogk)
Space complexity: O(n)

## C++

You have k servers numbered from 0 to k-1 that are being used to handle multiple requests simultaneously. Each server has infinite computational capacity but cannot handle more than one request at a time. The requests are assigned to servers according to a specific algorithm:

• The ith (0-indexed) request arrives.
• If all servers are busy, the request is dropped (not handled at all).
• If the (i % k)th server is available, assign the request to that server.
• Otherwise, assign the request to the next available server (wrapping around the list of servers and starting from 0 if necessary). For example, if the ith server is busy, try to assign the request to the (i+1)th server, then the (i+2)th server, and so on.

You are given a strictly increasing array arrival of positive integers, where arrival[i] represents the arrival time of the ith request, and another array load, where load[i] represents the load of the ith request (the time it takes to complete). Your goal is to find the busiest server(s). A server is considered busiest if it handled the most number of requests successfully among all the servers.

Return a list containing the IDs (0-indexed) of the busiest server(s). You may return the IDs in any order.

Example 1:

Input: k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3]
Output: 
Explanation:
All of the servers start out available.
The first 3 requests are handled by the first 3 servers in order.
Request 3 comes in. Server 0 is busy, so it's assigned to the next available server, which is 1.
Request 4 comes in. It cannot be handled since all servers are busy, so it is dropped.
Servers 0 and 2 handled one request each, while server 1 handled two requests. Hence server 1 is the busiest server.


Example 2:

Input: k = 3, arrival = [1,2,3,4], load = [1,2,1,2]
Output: 
Explanation:
The first 3 requests are handled by first 3 servers.
Request 3 comes in. It is handled by server 0 since the server is available.
Server 0 handled two requests, while servers 1 and 2 handled one request each. Hence server 0 is the busiest server.


Example 3:

Input: k = 3, arrival = [1,2,3], load = [10,12,11]
Output: [0,1,2]
Explanation: Each server handles a single request, so they are all considered the busiest.


Example 4:

Input: k = 3, arrival = [1,2,3,4,8,9,10], load = [5,2,10,3,1,2,2]
Output: 


Example 5:

Input: k = 1, arrival = , load = 
Output: 


Constraints:

• 1 <= k <= 105
• 1 <= arrival.length, load.length <= 105
• arrival.length == load.length
• 1 <= arrival[i], load[i] <= 109
• arrival is strictly increasing.

## Solution: Heap + TreeSet

Use a min heap to store the release time -> server.
Use a treeset to track the current available servers.
For reach request, check whether servers can be released at that time.

Time complexity: O(nlogk)
Space complexity: O(k)