Code Issues Solver

Minimum CPU Cores Calculator

This program uses a sweep line algorithm to determine the minimum number of CPU cores needed to execute overlapping processes defined by their start and end times, effectively managing core utilization in processor scheduling.

Prompt

Process scheduling algorithms are used by a CPU to optimally schedule the running processes. A core can execute one process at a time, but a CPU may have multiple cores.

There are n processes where the ith process starts its execution at start[i] and ends at end[i], both inclusive. Find the minimum number of cores required to execute the processes.

Example

n=3

start = [1, 3, 4]

end = [3, 5, 6].

If the CPU has only one core, the first process starts at 1 and erlds at 3. The second process starts at 3. Since both processes need a processor at 3, they overlap. There must be more than 1 core.

If the CPU has two cores, the first process runs on the first core from 1 to 3, the second runs on the second core from 3 to 5, and the third process runs on the first core from 4 to 6.

Return 2, the minimum number of cores required.Function Description

Complete the function getMinCores in the editor below.

ALL

getMinCores takes the following arguments:

int start[n]: the start times of processes

int end[n]: the end times of processes

Returns

int: the minimum number of cores required

Constraints

• 15n≤105

1 ≤ start[i] send[i] ≤ 109

► Input Format For Custom Testing▼ Sample Case 0

Sample Input For Custom Testing

STDIN

FUNCTION

→ start[] size n = 3

→

start = [1, 2, 3]

→ end[] size n = 3

→ end [3, 3, 5]

Sample Output

Explanation

Using 2 cores, the first and second processes finish at time 3. The third process starts at 3. Given the conflict, there must be at least 1 more core.

► Sample Case 1 Give the code for below syntax 1>#!/bin/python3...

Complete the 'getMinCores' function below.

13 #

The function is expected to return an INTEGER.

The function accepts following parameters:

#1. INTEGER_ARRAY start

#2. INTEGER_ARRAY end

def getMinCores (start, end):

Write your code here

23 > if name == 1 main :Here is the correct code to pass all the test cases on HackerRank:

import heapq

def getMinCores(start, end):
    # Combine start and end times into a list of tuples
        times = [(s, 1) for s in start] + [(e, -1) for e in end]
            
                # Sort the times
                    times.sort()
                        
                            # Initialize the minimum cores and current cores
                                min_cores = 0
                                    curr_cores = 0
                                        
                                            # Use a priority queue to keep track of the end times
                                                pq = []
                                                    
                                                        for time, delta in times:
                                                                # Remove the end times that are less than or equal to the current time
                                                                        while pq and pq[0] <= time:
                                                                                    curr_cores -= 1
                                                                                                heapq.heappop(pq)
                                                                                                        
                                                                                                                # Add the current time to the priority queue if it's a start time
                                                                                                                        if delta == 1:
                                                                                                                                    curr_cores += 1
                                                                                                                                                heapq.heappush(pq, time)
                                                                                                                                                        
                                                                                                                                                                # Update the minimum cores
                                                                                                                                                                        min_cores = max(min_cores, curr_cores)
                                                                                                                                                                            
                                                                                                                                                                                return min_cores

                                                                                                                                                                                if __name__ == '__main__':
                                                                                                                                                                                    # Test the function
                                                                                                                                                                                        start = [1, 2, 3]
                                                                                                                                                                                            end = [3, 3, 5]
                                                                                                                                                                                                print(getMinCores(start, end))  # Output: 3
                                                                                                                                                                                                ```

                                                                                                                                                                                                This code uses a priority queue to keep track of the end times, which allows it to efficiently remove the end times that are less than or equal to the current time. This ensures that the code passes all the test cases on HackerRank.def getMinCores(start, end):
                                                                                                                                                                                                    # Create events for start and end times
                                                                                                                                                                                                        events = []
                                                                                                                                                                                                            
                                                                                                                                                                                                                # Start event is +1, end event is -1
                                                                                                                                                                                                                    for i in range(len(start)):
                                                                                                                                                                                                                            events.append((start[i], 1))  # process start
                                                                                                                                                                                                                                    events.append((end[i] + 1, -1))  # process end

                                                                                                                                                                                                                                        # Sort events by time; if times are the same, end events (-1) come before start events (+1)
                                                                                                                                                                                                                                            events.sort(key=lambda x: (x[0], x[1]))
                                                                                                                                                                                                                                                
                                                                                                                                                                                                                                                    max_cores = 0
                                                                                                                                                                                                                                                        current_cores = 0
                                                                                                                                                                                                                                                            
                                                                                                                                                                                                                                                                # Sweep line algorithm
                                                                                                                                                                                                                                                                    for event in events:
                                                                                                                                                                                                                                                                            current_cores += event[1]
                                                                                                                                                                                                                                                                                    max_cores = max(max_cores, current_cores)
                                                                                                                                                                                                                                                                                        
                                                                                                                                                                                                                                                                                            return max_cores

                                                                                                                                                                                                                                                                                            # Example usage:
                                                                                                                                                                                                                                                                                            n = 3
                                                                                                                                                                                                                                                                                            start = [1, 3, 4]
                                                                                                                                                                                                                                                                                            end = [3, 5, 6]

                                                                                                                                                                                                                                                                                            print(getMinCores(start, end))  # Output: 2

Answer

Process Scheduling Algorithm: Minimum Cores Required

Problem Analysis

The problem at hand involves determining the minimum number of CPU cores necessary to execute a set of processes that have defined start and end times. The challenge arises from the overlap of the execution times of these processes.

Solution Approach

To address this, we can utilize a sweep line algorithm where we treat both the start and end times as events and track the number of active processes at any given time. The core idea involves:

Event Creation: Create events for each process' start and end time.
Event Sorting: Sort these events, prioritizing end events before start events if they occur at the same time.
Core Counting: Use a counter to track the current number of cores needed as we process each event sequentially.

Key Considerations:

Start events will increase the core requirement.
End events will decrease the core requirement.
The maximum count during this process will yield the minimum required cores.

Implementation

Here’s how to implement the aforementioned strategy in Python:

def getMinCores(start, end):
    # Create events for start (+1) and end (-1) times.
    events = []
    
    for i in range(len(start)):
        events.append((start[i], 1))  # Process start
        events.append((end[i] + 1, -1))  # Process end (exclusive end)

    # Sort events; end events should come before start in case of a tie.
    events.sort(key=lambda x: (x[0], x[1]))
    
    max_cores = 0
    current_cores = 0
    
    # Sweep line algorithm to count the cores required.
    for event in events:
        current_cores += event[1]  # Update current cores
        max_cores = max(max_cores, current_cores)  # Keep track of maximum cores needed
    
    return max_cores

# Example usage:
if __name__ == '__main__':
    start = [1, 3, 4]
    end = [3, 5, 6]
    print(getMinCores(start, end))  # Output: 2

Code Explanation:

Event Creation: For each process, we create and append two tuples: one for the start time and one for the end time incremented by one (to denote that the process is not running after end).
Sorting: Sorting is done to ensure that in case of simultaneous events, the end event takes precedence.
Counting Cores: A loop iterates over the sorted events and updates the count of current cores based on whether it’s a start (+1) or end (-1) event, simultaneously tracking the maximum core usage.

Complexity:

Time Complexity: O(n log n) due to sorting.
Space Complexity: O(n) for the events array.

Conclusion

This approach efficiently calculates the minimum number of CPU cores required to run overlapping processes while adhering to constraints and ensuring performance considerations. The provided code is ready for testing and integration into larger systems.