Problem Description

Imagine attending a party with a never-ending row of seats, where each friend chooses the first available seat to sit down. When a friend leaves, their seat becomes available for the next person who shows up. Everyone arrives and leaves the party at specific times. For instance, if friend 0 occupies chairs 0, 1, and 5, the next arriving friend will take seat 2. Now, we have a list detailing the arrival and departure times for each friend, and our goal is to find out which chair a particular friend (referred to as targetFriend) will occupy.

Intuition

To determine which seat targetFriend takes, we need to:

1. Keep track of all available chairs and occupied chairs at any given time.
2. Assign the lowest available chair number to each friend on arrival.
3. Release the chair when a friend leaves, making it available for new arrivals.

We can achieve this using a min-heap to represent the available chairs, ensuring we always get the lowest numbered chair. Another min-heap is used to manage occupied chairs, where the key is the leaving time. We keep track of friends' arriving and leaving times, and we check the availability of chairs at each arrival. When a friend leaves, we release their chair and put it back into the heap of available chairs.

When the targetFriend arrives, we extract the lowest available chair number from the heap and return it, as this is the chair they will occupy.

The process looks like this:

1. Initialize a min-heap with all possible chair numbers.
2. Sort the list of friends by arrival time so that we assign chairs in the order friends arrive.
3. Initialize a second min-heap to manage occupied chairs, sorted by leaving time.
4. Iterate over the sorted list, and at each friend's arrival:
   • Release the chairs of friends who have left (leaving time is less than or equal to the current arrival time).
   • Extract the next available chair from the min-heap of available chairs.
   • If the current friend is targetFriend, return the chosen chair number.
   • Add the current friend's chair and leaving time to the min-heap of occupied chairs.

By utilizing two heaps, we efficiently manage the incoming and outgoing friends and their chair selections in a way that fulfills all the constraints of the given problem.

Learn more about Heap (Priority Queue) patterns.

Solution Approach

The implementation of the solution is centered around efficiently tracking and managing available and occupied chairs using heaps. Here's how the provided solution works:

1. Initialization of the Min-Heaps:

   • We create a min-heap h representing all possible chair numbers. Initially, this heap contains all numbers from 0 to n-1, because there are n friends and hence at most n chairs that could be occupied at once.
   • We use the heapq library's heapify method to transform the list of chair numbers into a min-heap.
   • A second min-heap busy is used to track occupied chairs, with leaving times as keys, so we know when chairs become available.

2. Annotating Times with Indices:

   • We append each friend's index to their respective [arrival, leaving] times to keep a mapping of which friend is associated with which time slot.

3. Sorting Based on Arrival Times:

   • We then sort the times array by arrival times. Sorting is necessary to simulate the chronological order of events as friends arrive and depart.

4. Iterating Over Sorted Times:

   • We loop through each friend's arrival (a) and departure (b) times (after annotating them with the friend's index i):
     • For each arriving friend, we check if there are any occupied chairs that have become available (i.e., the current time is greater than or equal to the leaving time of some friends). This is done by checking the busy heap.
     • If there are such chairs, we push them back to the h heap of available chairs.

5. Allocating Chairs & Handling targetFriend:

   • We pop the lowest number from the h heap which represents the currently available chair with the smallest number.
   • If the index i of the currently arriving friend is equal to targetFriend, we return the chair number immediately since this is the chair targetFriend will sit on.
   • If not, we push the friend's chair along with their leaving time to the busy heap, indicating that this chair is now occupied until the leaving time.

6. Returning the Chair Number:

   • The loop continues until targetFriend is encountered. When targetFriend arrives, their chair number is returned.

The algorithm effectively uses min-heaps to ensure that we always allocate the smallest available chair number and free chairs up in the order people leave. By maintaining two separate heaps—one for available chairs sorted by chair number and another for busy (occupied) chairs sorted by leaving times—we ensure an efficient and orderly management of arriving and departing friends.

Notice that the solution has a fail-safe return value of -1, which is a pattern often used in functions to indicate that for some reason, no valid result was computed. However, in the context of this problem, this line is never reached because it is guaranteed that targetFriend will eventually arrive and be assigned a chair.

Example Walkthrough

Let's illustrate the solution approach with a simple example. Suppose we have 4 friends with the following arrival and departure times and we want to find the chair number for targetFriend identified by index 2:

• Friend 0: Arrives at 1, leaves at 4
• Friend 1: Arrives at 2, leaves at 5
• Friend 2: Arrives at 6, leaves at 9 (targetFriend)
• Friend 3: Arrives at 3, leaves at 7

1. Initialization of the Min-Heaps: We begin by initializing a min-heap h with chair numbers [0, 1, 2, 3]. The busy min-heap is initialized and empty because no friends have arrived yet.

2. Annotating Times with Indices: We modify the times list to include each friend's index:

   • Friend 0: [1, 4, 0]
   • Friend 1: [2, 5, 1]
   • Friend 2: [6, 9, 2] (targetFriend)
   • Friend 3: [3, 7, 3]

3. Sorting Based on Arrival Times: We sort the list based on arrival times:

   • Friend 0: [1, 4, 0]
   • Friend 1: [2, 5, 1]
   • Friend 3: [3, 7, 3]
   • Friend 2: [6, 9, 2] (targetFriend)

4. Iterating Over Sorted Times:

   • At time 1, friend 0 arrives. We pop 0 from h, and push [4, 0] onto busy.
   • At time 2, friend 1 arrives. We pop 1 from h, and push [5, 1] onto busy.
   • At time 3, friend 3 arrives. We pop 2 from h, and push [7, 2] onto busy.
   • Time 4 arrives, friend 0's leaving time. We pop [4, 0] from busy and push 0 back onto h.
   • We continue like this until all friends have arrived and left or until the targetFriend arrives.

5. Allocating Chairs & Handling targetFriend:

   • When targetFriend (index 2) arrives at time 6, friends 0 and 1 have left, freeing chairs 0 and 1. Since 0 is the lowest number in h, we pop it and assign it to the index 2 friend.
   • We return the chair number 0 for the targetFriend because they are the next to arrive.

This example navigates through the arrival and departure of friends, effectively managing chair allocations using two heaps and efficiently identifies the chair number for the targetFriend, which is 0 in this scenario.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code can be analyzed as follows:

1. The initial heapification of h which is a list of n different numbers has a time complexity of O(n).

2. Sorting the times list, which has n elements, has a time complexity of O(n log n).

3. The for loop iterates over each of the n elements of the times list. Inside the loop, operations related to heap operations (heappop and heappush) are performed. In the worst case, each operation can take O(log n).

The while loop will pop elements out of the busy heap and push them back into the h heap. Each of these operations takes O(log n). However, each chair becomes free at most once during the whole process, so the total number of heappush and heappop operations for the busy list will be O(n) across the entire runtime of the program.

Finally, we have the heappop from h and conditional heappush in the busy heap inside the for loop. Since this happens once per guest (n times in total), we can represent these operations as n * O(log n).

Therefore, the overall time complexity of the loop is O(n log n).

4. Taking both sorting and heap operations into account, the overall time complexity of the entire function can be represented as O(n log n) because the sort dominates the O(n) initialization of the h heap.

Space Complexity

The space complexity of the provided code can be analyzed as follows:

1. Auxiliary space for the h heap which stores at most n elements is O(n).

2. The busy heap also, in the worst case, can grow up to O(n). However, because the heaps do not grow independently (when an element is removed from busy, it is added to h, and vice versa) and are bounded by the number of chairs n, we still represent the total space used by both heaps as O(n).

3. We also made modifications to the times array, adding the index i, but this is still within the O(n) space usage because each sub-list is only extended by one element.

Therefore, the overall space complexity is O(n) due to the heap structures and times list modifications.

Learn more about how to find time and space complexity quickly using problem constraints.