Maximum Number of Events That Can Be Attended II

Hard

Apple

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Topics:

ArraysBinary SearchDynamic Programming

You are given an array of events where events[i] = [startDay_i, endDay_i, value_i]. The ith event starts at startDay_i and ends at endDay_i, and if you attend this event, you will receive a value of value_i. You are also given an integer k which represents the maximum number of events you can attend.

You can only attend one event at a time. If you choose to attend an event, you must attend the entire event. Note that the end day is inclusive: that is, you cannot attend two events where one of them starts and the other ends on the same day.

Return the maximum sum of values that you can receive by attending events.

Example 1:

events = [[1,2,4],[3,4,3],[2,3,1]], k = 2

Output: 7

Explanation: Choose the green events, 0 and 1 (0-indexed) for a total value of 4 + 3 = 7.

Example 2:

events = [[1,2,4],[3,4,3],[2,3,10]], k = 2

Output: 10

Explanation: Choose event 2 for a total value of 10. Notice that you cannot attend any other event as they overlap, and that you do not have to attend k events.

Example 3:

events = [[1,1,1],[2,2,2],[3,3,3],[4,4,4]], k = 3

Output: 9

Explanation: Although the events do not overlap, you can only attend 3 events. Pick the highest valued three.

How would you solve this problem efficiently?

Maximum Number of Events That Can Be Attended II

Hard

Apple

3 views

Topics:

ArraysBinary SearchDynamic Programming

You are given an array of events where events[i] = [startDay_i, endDay_i, value_i]. The ith event starts at startDay_i and ends at endDay_i, and if you attend this event, you will receive a value of value_i. You are also given an integer k which represents the maximum number of events you can attend.

You can only attend one event at a time. If you choose to attend an event, you must attend the entire event. Note that the end day is inclusive: that is, you cannot attend two events where one of them starts and the other ends on the same day.

Return the maximum sum of values that you can receive by attending events.

Example 1:

events = [[1,2,4],[3,4,3],[2,3,1]], k = 2

Output: 7

Explanation: Choose the green events, 0 and 1 (0-indexed) for a total value of 4 + 3 = 7.

Example 2:

events = [[1,2,4],[3,4,3],[2,3,10]], k = 2

Output: 10

Explanation: Choose event 2 for a total value of 10. Notice that you cannot attend any other event as they overlap, and that you do not have to attend k events.

Example 3:

events = [[1,1,1],[2,2,2],[3,3,3],[4,4,4]], k = 3

Output: 9

Explanation: Although the events do not overlap, you can only attend 3 events. Pick the highest valued three.

How would you solve this problem efficiently?

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the number of events, and the values of startDay, endDay, and value? Are they all positive integers?
If it's impossible to attend any events given the constraints, what should I return? Should I return 0?
Are the events sorted by startDay, endDay, or value, or are they unsorted?
Can events overlap in time? How should I handle overlapping events with respect to the limit 'k'?
Is 'k' guaranteed to be less than or equal to the number of events, and what is the maximum value for 'k'?

Brute Force Solution

Approach

The brute force approach to maximizing attended events involves exploring every possible combination of events. We'll consider attending or skipping each event to find the best possible set of events without exceeding the allowed event limit. By meticulously checking each combination, we guarantee we'll find the optimal solution.

Here's how the algorithm would work step-by-step:

Start with the first event and consider two possibilities: either attend it or skip it.
If you attend the first event, mark it as attended and reduce the number of allowed events by one. Move to the next event.
If you skip the first event, don't mark it as attended and don't reduce the number of allowed events. Move to the next event.
Repeat this process (attend or skip) for every event, making sure you never try to attend more events than you are allowed.
Each time you reach the end of the event list, you have explored one possible combination of attended events.
Calculate the total 'value' (reward) of the attended events in this combination.
Keep track of the combination that gave you the highest total 'value' of attended events.
Once you have explored every single combination of attending or skipping events, the combination with the highest 'value' is your answer.

Code Implementation

def max_value_of_events_brute_force(events, max_events_allowed):
    max_total_value = 0

    def explore_combinations(current_index, events_attended, current_value):
        nonlocal max_total_value

        # If we've considered all events, update the maximum value.
        if current_index == len(events):
            max_total_value = max(max_total_value, current_value)
            return

        # If we've reached the maximum number of events, return.
        if events_attended == max_events_allowed:
            max_total_value = max(max_total_value, current_value)
            return

        # Explore the possibility of skipping the current event.
        explore_combinations(current_index + 1, events_attended,
                             current_value)

        # Explore the possibility of attending the current event.
        # Only proceed if we haven't exceeded the allowed limit
        explore_combinations(current_index + 1, events_attended + 1,
                             current_value + events[current_index][2])

    explore_combinations(0, 0, 0)
    return max_total_value

Big(O) Analysis

Time Complexity

O(2^n) – The provided brute force approach explores every possible combination of attending or skipping each event. Given 'n' events, each event has two choices: attend or skip. Therefore, the total number of possible combinations is 2 * 2 * ... * 2 (n times), which equals 2^n. For each combination, we need to calculate the total value of attended events, which takes O(n) time in the worst case. However, the dominant factor is the number of combinations, leading to a time complexity of O(2^n).

Space Complexity

O(1) – The brute force approach described explores all combinations using a recursive or iterative method, but the provided explanation doesn't explicitly mention storing intermediate combinations. The core logic only involves keeping track of the current event being considered and the remaining allowed events. Without any mentioned auxiliary data structures for storing combinations or results, the space used consists mainly of a few variables for tracking the current event, the number of allowed events, and potentially a variable for the maximum value found so far, which takes constant space regardless of the number of events N. Therefore, the space complexity is O(1).

Optimal Solution

Approach

The problem asks us to find the maximum value we can get by attending a limited number of events. The trick is to use a technique that avoids recomputing the same information over and over again by remembering the best choices we've made so far.

Here's how the algorithm would work step-by-step:

First, sort the events by their end date so we can process them in chronological order.
Think of building a table where each cell represents the maximum value we can obtain by considering the first few events and attending up to a certain number of events.
Start filling in the table from the top left. Each cell's value is determined by making a choice: either we attend the current event or we don't.
If we don't attend the current event, the value is simply the maximum value from the previous row (considering one less event).
If we do attend the current event, we need to find the latest event that finishes before the current event starts. This is important, as we cannot overlap events.
If we attend the current event, the value is the sum of the current event's value plus the maximum value obtained from attending events before the non-overlapping event we identified.
We choose the bigger of the two values (attending or not attending the current event) and put it in the table cell.
After filling in the entire table, the bottom right cell will contain the maximum value we can obtain by attending up to the given limit of events.

Code Implementation

def max_value(events, max_events_to_attend):
    events.sort(key=lambda x: x[1])
    number_of_events = len(events)

    # dp[i][j] is max value using first i events and attending at most j events
    dp_table = [[0] * (max_events_to_attend + 1) for _ in range(number_of_events + 1)]

    for event_index in range(1, number_of_events + 1):
        start_time, end_time, value = events[event_index - 1]
        for events_attended in range(1, max_events_to_attend + 1):

            # Option 1: Don't attend the current event
            dp_table[event_index][events_attended] = dp_table[event_index - 1][events_attended]

            # Find the latest non-overlapping event
            latest_non_overlapping_event = -1
            for previous_event_index in range(event_index - 1, 0 -1, -1):
                if events[previous_event_index][1] < start_time:
                    latest_non_overlapping_event = previous_event_index + 1 -1
                    break

            # Option 2: Attend the current event if it doesn't exceed max events
            if latest_non_overlapping_event == -1:
                value_from_previous_events = 0
            else:
                value_from_previous_events = dp_table[latest_non_overlapping_event + 1][events_attended - 1]

            # Choose the max value between attending or not attending
            dp_table[event_index][events_attended] = max(dp_table[event_index][events_attended], \
                                                  value_from_previous_events + value)

    # Result is the max value attending up to max_events_to_attend events.
    return dp_table[number_of_events][max_events_to_attend]

Big(O) Analysis

Time Complexity

O(n * k * log n) – The events are sorted by their end dates, which takes O(n log n) time. Then, a table of size n * k is constructed where n is the number of events and k is the maximum number of events that can be attended. Filling each cell of this table involves a binary search (log n) to find the latest non-overlapping event. Since each of the n * k cells requires a binary search, the overall time complexity is O(n * k * log n). The sorting cost O(n log n) is dominated by O(n * k * log n).

Space Complexity

O(N*K) – The described solution constructs a table (a 2D array) to store intermediate results. The dimensions of this table are determined by the number of events (N) and the maximum number of events that can be attended (K). Therefore, the table will have N rows and K columns, leading to a space requirement proportional to N*K. This table is the dominant factor in the auxiliary space used by the algorithm. Thus, the space complexity is O(N*K).

Edge Cases

Case	How to Handle
Empty events array	Return 0 as no events can be attended.
k = 0 (no events allowed)	Return 0 since no events can be selected.
Events with the same start day	Sort events primarily by start day and secondarily by end day to process events with earlier start times first.
Events with the same end day	When events have the same end day, prefer the event with the higher value to maximize profit.
k is greater than or equal to the number of events	Iterate through all events to find maximum possible value.
Large event values that could cause integer overflow	Use 64-bit integers (long) to prevent potential overflows when summing up event values.
Very large number of events (close to constraint limit)	Ensure dynamic programming table is efficiently allocated to avoid excessive memory usage or out-of-memory errors.
Events with very close start and end times (duration of 1)	The solution should handle such events without any issues as these are valid events.