Single-Threaded CPU Python Solution with Tests

lcpy gen -n 1834   # by problem number
lcpy gen -s single_threaded_cpu   # by problem name

Input: tasks = [[1,2],[2,4],[3,2],[4,1]]
Output: [0,2,3,1]
Explanation: The events go as follows:
- At time = 1, task 0 is available to process. Available tasks = {0}.
- Also at time = 1, the idle CPU starts processing task 0. Available tasks = {}.
- At time = 2, task 1 is available to process. Available tasks = {1}.
- At time = 3, task 2 is available to process. Available tasks = {1, 2}.
- Also at time = 3, the CPU finishes task 0 and starts processing task 2 as it is the shortest. Available tasks = {1}.
- At time = 4, task 3 is available to process. Available tasks = {1, 3}.
- At time = 5, the CPU finishes task 2 and starts processing task 3 as it is the shortest. Available tasks = {1}.
- At time = 6, the CPU finishes task 3 and starts processing task 1. Available tasks = {}.
- At time = 10, the CPU finishes task 1 and becomes idle.

Input: tasks = [[7,10],[7,12],[7,5],[7,4],[7,2]]
Output: [4,3,2,0,1]
Explanation: The events go as follows:
- At time = 7, all the tasks become available. Available tasks = {0,1,2,3,4}.
- Also at time = 7, the idle CPU starts processing task 4. Available tasks = {0,1,2,3}.
- At time = 9, the CPU finishes task 4 and starts processing task 3. Available tasks = {0,1,2}.
- At time = 13, the CPU finishes task 3 and starts processing task 2. Available tasks = {0,1}.
- At time = 18, the CPU finishes task 2 and starts processing task 0. Available tasks = {1}.
- At time = 28, the CPU finishes task 0 and starts processing task 1. Available tasks = {}.
- At time = 40, the CPU finishes task 1 and becomes idle.

import heapq


class Solution:
    # Time: O(n log n)
    # Space: O(n)
    def get_order(self, tasks: list[list[int]]) -> list[int]:
        indexed = sorted(range(len(tasks)), key=lambda i: tasks[i][0])

        order: list[int] = []
        heap: list[tuple[int, int]] = []
        time = 0
        pointer = 0
        n = len(tasks)

        while len(order) < n:
            while pointer < n and tasks[indexed[pointer]][0] <= time:
                idx = indexed[pointer]
                heapq.heappush(heap, (tasks[idx][1], idx))
                pointer += 1

            if heap:
                proc_time, idx = heapq.heappop(heap)
                time += proc_time
                order.append(idx)
            else:
                time = tasks[indexed[pointer]][0]

        return order