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Posts tagged as “heap”

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 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

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)

C++

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++

花花酱 LeetCode 1606. Find Servers That Handled Most Number of Requests

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: [1] 
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: [0]
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: [1]

Example 5:

Input: k = 1, arrival = [1], load = [1]
Output: [0]

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)

C++

花花酱 LeetCode 1488. Avoid Flood in The City

Your country has an infinite number of lakes. Initially, all the lakes are empty, but when it rains over the nth lake, the nth lake becomes full of water. If it rains over a lake which is full of water, there will be a flood. Your goal is to avoid the flood in any lake.

Given an integer array rains where:

  • rains[i] > 0 means there will be rains over the rains[i] lake.
  • rains[i] == 0 means there are no rains this day and you can choose one lake this day and dry it.

Return an array ans where:

  • ans.length == rains.length
  • ans[i] == -1 if rains[i] > 0.
  • ans[i] is the lake you choose to dry in the ith day if rains[i] == 0.

If there are multiple valid answers return any of them. If it is impossible to avoid flood return an empty array.

Notice that if you chose to dry a full lake, it becomes empty, but if you chose to dry an empty lake, nothing changes. (see example 4)

Example 1:

Input: rains = [1,2,3,4]
Output: [-1,-1,-1,-1]
Explanation: After the first day full lakes are [1]
After the second day full lakes are [1,2]
After the third day full lakes are [1,2,3]
After the fourth day full lakes are [1,2,3,4]
There's no day to dry any lake and there is no flood in any lake.

Example 2:

Input: rains = [1,2,0,0,2,1]
Output: [-1,-1,2,1,-1,-1]
Explanation: After the first day full lakes are [1]
After the second day full lakes are [1,2]
After the third day, we dry lake 2. Full lakes are [1]
After the fourth day, we dry lake 1. There is no full lakes.
After the fifth day, full lakes are [2].
After the sixth day, full lakes are [1,2].
It is easy that this scenario is flood-free. [-1,-1,1,2,-1,-1] is another acceptable scenario.

Example 3:

Input: rains = [1,2,0,1,2]
Output: []
Explanation: After the second day, full lakes are  [1,2]. We have to dry one lake in the third day.
After that, it will rain over lakes [1,2]. It's easy to prove that no matter which lake you choose to dry in the 3rd day, the other one will flood.

Example 4:

Input: rains = [69,0,0,0,69]
Output: [-1,69,1,1,-1]
Explanation: Any solution on one of the forms [-1,69,x,y,-1], [-1,x,69,y,-1] or [-1,x,y,69,-1] is acceptable where 1 <= x,y <= 10^9

Example 5:

Input: rains = [10,20,20]
Output: []
Explanation: It will rain over lake 20 two consecutive days. There is no chance to dry any lake.

Constraints:

  • 1 <= rains.length <= 10^5
  • 0 <= rains[i] <= 10^9

Solution: Binary Search

Store the days we can dry a lake in a treeset.
Store the last day when a lake becomes full in a hashtable.
Whenever we encounter a full lake, try to find the first available day that we can dry it. If no such day, return no answer.

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

C++

花花酱 LeetCode 1094. Car Pooling

You are driving a vehicle that has capacity empty seats initially available for passengers.  The vehicle only drives east (ie. it cannot turn around and drive west.)

Given a list of tripstrip[i] = [num_passengers, start_location, end_location] contains information about the i-th trip: the number of passengers that must be picked up, and the locations to pick them up and drop them off.  The locations are given as the number of kilometers due east from your vehicle’s initial location.

Return true if and only if it is possible to pick up and drop off all passengers for all the given trips. 

Example 1:

Input: trips = [[2,1,5],[3,3,7]], capacity = 4
Output: false

Example 2:

Input: trips = [[2,1,5],[3,3,7]], capacity = 5
Output: true

Example 3:

Input: trips = [[2,1,5],[3,5,7]], capacity = 3
Output: true

Example 4:

Input: trips = [[3,2,7],[3,7,9],[8,3,9]], capacity = 11
Output: true

Solution1: Min heap

Sort events by location

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

C++

Solution 2: Preprocessing

Time complexity: O(n)
Space complexity: O(1000)

C++