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Posts published in “Priority Queue”

花花酱 LeetCode 1851. Minimum Interval to Include Each Query

You are given a 2D integer array intervals, where intervals[i] = [lefti, righti] describes the ith interval starting at lefti and ending at righti (inclusive). The size of an interval is defined as the number of integers it contains, or more formally righti - lefti + 1.

You are also given an integer array queries. The answer to the jth query is the size of the smallest interval i such that lefti <= queries[j] <= righti. If no such interval exists, the answer is -1.

Return an array containing the answers to the queries.

Example 1:

Input: intervals = [[1,4],[2,4],[3,6],[4,4]], queries = [2,3,4,5]
Output: [3,3,1,4]
Explanation: The queries are processed as follows:
- Query = 2: The interval [2,4] is the smallest interval containing 2. The answer is 4 - 2 + 1 = 3.
- Query = 3: The interval [2,4] is the smallest interval containing 3. The answer is 4 - 2 + 1 = 3.
- Query = 4: The interval [4,4] is the smallest interval containing 4. The answer is 4 - 4 + 1 = 1.
- Query = 5: The interval [3,6] is the smallest interval containing 5. The answer is 6 - 3 + 1 = 4.

Example 2:

Input: intervals = [[2,3],[2,5],[1,8],[20,25]], queries = [2,19,5,22]
Output: [2,-1,4,6]
Explanation: The queries are processed as follows:
- Query = 2: The interval [2,3] is the smallest interval containing 2. The answer is 3 - 2 + 1 = 2.
- Query = 19: None of the intervals contain 19. The answer is -1.
- Query = 5: The interval [2,5] is the smallest interval containing 5. The answer is 5 - 2 + 1 = 4.
- Query = 22: The interval [20,25] is the smallest interval containing 22. The answer is 25 - 20 + 1 = 6.

Constraints:

  • 1 <= intervals.length <= 105
  • 1 <= queries.length <= 105
  • intervals[i].length == 2
  • 1 <= lefti <= righti <= 107
  • 1 <= queries[j] <= 107

Solution: Offline Processing + Priority Queue

Similar to 花花酱 LeetCode 1847. Closest Room

Sort intervals by right in descending order, sort queries in descending. Add valid intervals into the priority queue (or treeset) ordered by size in ascending order. Erase invalid ones. The first one (if any) will be the one with the smallest size that contains the current query.

Time complexity: O(nlogn + mlogm + mlogn)
Space complexity: O(m + n)

C++

LeetCode 1801. Number of Orders in the Backlog

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

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

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

C++

花花酱 LeetCode 1792. Maximum Average Pass Ratio

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)

C++

Python3

花花酱 LeetCode 1705. Maximum Number of Eaten Apples

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++

花花酱 LeetCode 1675. Minimize Deviation in Array

You are given an array nums of n positive integers.

You can perform two types of operations on any element of the array any number of times:

  • If the element is evendivide it by 2.
    • For example, if the array is [1,2,3,4], then you can do this operation on the last element, and the array will be [1,2,3,2].
  • If the element is oddmultiply it by 2.
    • For example, if the array is [1,2,3,4], then you can do this operation on the first element, and the array will be [2,2,3,4].

The deviation of the array is the maximum difference between any two elements in the array.

Return the minimum deviation the array can have after performing some number of operations.

Example 1:

Input: nums = [1,2,3,4]
Output: 1
Explanation: You can transform the array to [1,2,3,2], then to [2,2,3,2], then the deviation will be 3 - 2 = 1.

Example 2:

Input: nums = [4,1,5,20,3]
Output: 3
Explanation: You can transform the array after two operations to [4,2,5,5,3], then the deviation will be 5 - 2 = 3.

Example 3:

Input: nums = [2,10,8]
Output: 3

Constraints:

  • n == nums.length
  • 2 <= n <= 105
  • 1 <= nums[i] <= 109

Solution: Priority Queue

If we double an odd number it becomes an even number, then we can only divide it by two which gives us back the original number. So we can pre-double all the odd numbers and only do division in the following process.

We push all numbers including pre-doubled odd ones onto a priority queue, and track the difference between the largest and smallest number.

Each time, we pop the largest number out and divide it by two then put it back to the priority queue, until the largest number becomes odd. We can not discard it and divide any other smaller numbers by two will only increase the max difference, so we can stop here.

ex1: [3, 5, 8] => [6, 8, 10] (pre-double) => [5, 6, 8] => [4, 5, 6] => [3, 4, 5] max diff is 5 – 3 = 2
ex2: [4,1,5,20,3] => [2, 4, 6, 10, 20] (pre-double) => [2, 4, 6, 10] => [2, 4, 5, 6] => [2,3,4,5] max diff = 5-2 = 3

Time complexity: O(n*logm*logn)
Space complexity: O(n)

C++/Set

C++/PQ