Two Best Non-Overlapping Events

class Solution {
public:
    int maxTwoEvents(vector<vector<int>>& eventList) {
        vector<array<int, 3>> intervals;
        for (auto& event : eventList) {
            intervals.push_back({event[0], 1, event[2]});  // Start of an event
            intervals.push_back({event[1] + 1, 0, event[2]});  // End of an event
        }
        int maxResult = 0, currentMax = 0;
        sort(intervals.begin(), intervals.end());
        for (auto& interval : intervals) {
            if (interval[1]) {  // Start event; calculate potential max result
                maxResult = max(maxResult, interval[2] + currentMax);
            } else {  // End event; update current max for ongoing calculations
                currentMax = max(currentMax, interval[2]);
            }
        }
        return maxResult;
    }
};

The solution provided for determining the two best non-overlapping events from a list aims to compute the maximum combined value of any two non-overlapping events. Written in C++, the code leverages sorting and the sweep line algorithm for efficient calculation. Below is a breakdown of how the code approaches the problem:

• First, it initializes a vector of arrays to store both start and endpoint information of events. This setup also includes the value of the event.
• Each event undergoes splitting into two entries:
  • The start time is marked by a '1' to signify starting a new event.
  • The end time is recorded as 'event end time + 1', marked with a '0' to indicate the event's conclusion.
• This vector of interval entries is then sorted based on time. If multiple events share the same time, start times are prioritized as they are marked by '1', as per the sort routine in C++.
• The algorithm iterates through the sorted interval entries, using a variable currentMax to track the maximum value found up to the end of the latest event.
• During the iteration:
  • If a start of an event is encountered, it computes a potential maximum by adding the value of the current event to currentMax and checks if this combination exceeds any previously recorded combination stored in maxResult.
  • When an event end is encountered, it updates currentMax to ensure it holds the maximum ending event value so far.
• The result, maxResult, holds the maximum value of any two non-overlapping events, which is returned at the end.

Overall, the approach ensures efficient management of large data sets by minimizing unnecessary comparisons, and by logically breaking down event times into start and end points, it provides a straightforward method to determine overlaps and maximize value calculation.

class Solution {

    public int maxSumOfTwoEvents(int[][] events) {
        List<int[]> eventTimes = new ArrayList<>();

        // Transform events into points of start time, end time, and their values
        for (int[] event : events) {
            eventTimes.add(new int[] { event[0], 1, event[2] }); // 1 indicates the start of an event
            eventTimes.add(new int[] { event[1] + 1, 0, event[2] }); // 0 indicates the end of an event
        }

        // Sort the event times by time point, and start times before end times if equal
        eventTimes.sort((x, y) ->
            x[0] != y[0]
                ? Integer.compare(x[0], y[0])
                : Integer.compare(x[1], y[1])
        );

        int maximumSum = 0, currentMaxValue = 0;

        // Iterate through each time point
        for (int[] time : eventTimes) {
            if (time[1] == 1) { // Handling start of an event
                maximumSum = Math.max(maximumSum, time[2] + currentMaxValue);
            } else {  // Handling end of an event
                currentMaxValue = Math.max(currentMaxValue, time[2]);
            }
        }

        return maximumSum;
    }
}

In this Java solution, the objective is to compute the maximum sum of two non-overlapping events' values from a given list of events. Each event is represented by an array [start, end, value]. The method named maxSumOfTwoEvents is used to achieve this by following the steps below:

1. Initialize an empty list of integer arrays named eventTimes to store transformed events.
2. Convert each event into two records: one marking the start and another marking the end. During this conversion:
   • An event's start is represented as [start_time, 1, value].
   • The end is adjusted to one unit after the actual end_time to clearly demarcate the scope of the event, represented as [end_time + 1, 0, value].
3. Sort the eventTimes list based on the event's time. If multiple events have the same time, starts are processed before ends.
4. Initialize two integer variables, maximumSum to track the maximum sum of values of two non-overlapping events, and currentMaxValue to capture the value of the most lucrative event encountered so far.
5. Iterate through each record in the sorted eventTimes list:
   • If the record marks the start of an event, calculate the potential maximum sum by adding the event's value to currentMaxValue. Update maximumSum if this calculated sum exceeds the current maximumSum.
   • If the record marks the end of an event, update currentMaxValue to be the maximum of itself and the event's value, to ensure it reflects the highest value of any standalone event so far.
6. Return the value of maximumSum, which now holds the maximum sum obtained from any two non-overlapping events.

This solution leverages sorting and linear iteration to efficiently determine the maximum possible value sum from any two non-overlapping events, ensuring that the overall time complexity remains manageable.

class Solution:
    def maximumSumOfTwoEvents(self, event_list):
        event_times = []
        for event in event_list:
            # Add start information
            event_times.append([event[0], 1, event[2]])
            # Add end information
            event_times.append([event[1] + 1, 0, event[2]])

        max_result, current_max = 0, 0
        event_times.sort()

        for item in event_times:
            # When processing start of an event
            if item[1]:
                max_result = max(max_result, item[2] + current_max)
            else:
                current_max = max(current_max, item[2])

        return max_result

The code provided is a Python solution designed for finding the maximum sum of values from two non-overlapping events in a given list. The approach utilized involves a series of steps to process and calculate the desired outcome:

• Initialization of the Event List:

  • Create an empty list event_times to store events' start, end, and value information distinctly.

• Populating Time Stamps:

  • Iterate through each event in the event_list to store both the start time (event[0]) and just after the end time (event[1] + 1) of an event along with its value (event[2]). This helps in marking the beginning and the point just beyond the ending of an event.

• Sorting Event Times:

  • Sort the event_times list to ensure that the events are processed in chronological order, which is necessary for the neat calculation of non-overlapping events.

• Finding Maximum Sum:

  • Iterate over the sorted event_times. Two distinct cases are considered for each time stamp:
    • Event Start (item[1] is True): Update max_result by considering the maximum between the current max_result or the sum of the event's value (item[2]) and current_max. This condition helps in checking the value if the current event can be added without overlapping with any previous event.
    • Event End (item[1] is False): Update current_max by comparing it with the event's value. This value represents the maximum sum of events processed up to this point without overlapping.

• Return Result:

  • At the end of the loop, max_result will store the maximum sum of two non-overlapping events, which is then returned.

This approach smartly uses sorting and linear traversal to maintain efficiency, while adequately handling non-overlapping conditions using logical separations of event starts and ends. The code is expected to be efficient with a complexity primarily dictated by the sort operation, i.e., O(n log n), where n is the number of events. The use of max operations within the loop does not significantly increase complexity but ensures the correct and optimal solution is derived by processing events in their orderly sequence. This method is both effective in tackling the problem of finding two best non-overlapping events and illustrating efficient use of data processing in Python.