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When multiple processes need to be executed by a CPU, it’s important to efficiently manage the core resources available to avoid overloading any single core. Given n processes, each with a start and end time, the task is to determine the minimum number of CPU cores required to execute all processes without overlapping.

In this problem, each process starts at start[i] and ends at end[i]. The goal is to calculate the minimum number of cores needed to handle these processes optimally.

Approach

Key Concept: Interval Overlap

The key to solving this problem is to track the maximum number of overlapping processes at any point in time. Each overlap means that an additional core is needed to handle the concurrent processes. The maximum number of overlapping processes will give us the minimum number of cores required to execute all processes without any conflicts.

Steps to Solve the Problem

1. Extract Events:

• For each process, there are two events: a start event and an end event.
• A start event indicates that a process has begun, requiring a core.
• An end event indicates that a process has finished, freeing up a core.

1. Sort Events:

• Sort all events by time. If two events occur at the same time, prioritize the end event before the start event to avoid unnecessary overlaps.

1. Simulate Core Allocation:

• Iterate through the sorted events while keeping track of the number of active processes (those that have started but not finished).
• Track the maximum number of active processes at any point, which represents the minimum number of cores required.

Java Code Implementation

Here’s how you can implement the above approach in Java:

java

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class CoreScheduling {

    public static int minCoresRequired(int[] start, int[] end) {
        // List to store all events
        List<Event> events = new ArrayList<>();

        // Add start and end events to the list
        for (int i = 0; i < start.length; i++) {
            events.add(new Event(start[i], true));  // Process starts
            events.add(new Event(end[i], false));   // Process ends
        }

        // Sort events: by time, then by type (end before start if times are the same)
        Collections.sort(events);

        int maxCores = 0;
        int currentCores = 0;

        // Simulate the events
        for (Event event : events) {
            if (event.isStart) {
                currentCores++; // A process starts, need an additional core
                maxCores = Math.max(maxCores, currentCores); // Track max cores in use
            } else {
                currentCores--; // A process ends, release a core
            }
        }

        return maxCores;
    }

    // Event class representing a start or end of a process
    static class Event implements Comparable<Event> {
        int time;
        boolean isStart;

        Event(int time, boolean isStart) {
            this.time = time;
            this.isStart = isStart;
        }

        @Override
        public int compareTo(Event other) {
            if (this.time == other.time) {
                return Boolean.compare(this.isStart, other.isStart);
            }
            return Integer.compare(this.time, other.time);
        }
    }

    public static void main(String[] args) {
        // Example usage
        int[] start = {1, 2, 3, 5};
        int[] end = {4, 5, 7, 8};

        int minCores = minCoresRequired(start, end);
        System.out.println("Minimum number of cores required: " + minCores); // Output: 2
    }
}

Explanation of the Code

Event Class:

• This class represents an event, either the start or end of a process. It includes:
• time: The time when the event occurs.
• isStart: A boolean flag to distinguish between start and end events.
• The compareTo method ensures that events are sorted by time, with end events prioritized when times are identical (to avoid unnecessary overlaps).

minCoresRequired Function:

1. Convert Start/End Times into Events:

• We create an event for each process’s start and end time.

1. Sort Events:

• The events are sorted by time. If two events occur at the same time, the end event is given priority to minimize core usage.

1. Simulate the Process Timeline:

• We iterate over the sorted events, updating the count of active cores and tracking the maximum number of cores used at any point in time.

1. Return the Result:

• The result is the maximum number of cores needed to run all processes concurrently.

Time and Space Complexity

Time Complexity:

• Sorting the events: O(nlog⁡n)O(n \log n)O(nlogn), where nnn is the number of processes.
• Iterating over the events: O(n)O(n)O(n), since we iterate over each event once.

Overall time complexity: O(nlog⁡n)O(n \log n)O(nlogn)

Space Complexity:

• Space for storing events: O(n)O(n)O(n), since we store two events for each process (start and end).

Overall space complexity: O(n)O(n)O(n)

Example Walkthrough

Let’s walk through the provided example:

java

int[] start = {1, 2, 3, 5}; // Start times
int[] end = {4, 5, 7, 8};   // End times

Process:

1. The list of events after conversion:

• (1, start), (4, end), (2, start), (5, end), (3, start), (7, end), (5, start), (8, end)

1. The events are sorted by time:

• (1, start), (2, start), (3, start), (4, end), (5, end), (5, start), (7, end), (8, end)

1. Iterating through the events:

• At time 1: A process starts → Active cores: 1
• At time 2: A process starts → Active cores: 2
• At time 3: A process starts → Active cores: 3
• At time 4: A process ends → Active cores: 2
• At time 5: A process ends → Active cores: 1
• At time 5: A process starts → Active cores: 2
• At time 7: A process ends → Active cores: 1
• At time 8: A process ends → Active cores: 0

1. The maximum number of cores used is 2, so the result is 2.

Output:

swift

Minimum number of cores required: 2

Conclusion

This approach leverages a greedy algorithm to efficiently calculate the minimum number of cores required for process scheduling. By focusing on the overlaps between processes and simulating the events, we ensure that the CPU is optimally scheduled, maximizing resource utilization while avoiding unnecessary core usage.