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* to add job seq program * to add job seq program * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * to add definitions in parameters * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * to add definitions in parameters * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * to add definitions in parameters * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * changes as recommended * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * type hint error resolved * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * removed lambda * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * import stmts order * Update and rename job_sequence.py to job_sequence_with_deadline.py --------- Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com> Co-authored-by: Christian Clauss <cclauss@me.com>
63 lines
1.8 KiB
Python
63 lines
1.8 KiB
Python
"""
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Given a list of tasks, each with a deadline and reward, calculate which tasks can be
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completed to yield the maximum reward. Each task takes one unit of time to complete,
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and we can only work on one task at a time. Once a task has passed its deadline, it
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can no longer be scheduled.
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Example :
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tasks_info = [(4, 20), (1, 10), (1, 40), (1, 30)]
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max_tasks will return (2, [2, 0]) -
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Scheduling these tasks would result in a reward of 40 + 20
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This problem can be solved using the concept of "GREEDY ALGORITHM".
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Time Complexity - O(n log n)
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https://medium.com/@nihardudhat2000/job-sequencing-with-deadline-17ddbb5890b5
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"""
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from dataclasses import dataclass
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from operator import attrgetter
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@dataclass
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class Task:
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task_id: int
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deadline: int
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reward: int
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def max_tasks(tasks_info: list[tuple[int, int]]) -> list[int]:
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"""
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Create a list of Task objects that are sorted so the highest rewards come first.
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Return a list of those task ids that can be completed before i becomes too high.
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>>> max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)])
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[2, 0]
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>>> max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)])
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[3, 2]
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>>> max_tasks([(9, 10)])
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[0]
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>>> max_tasks([(-9, 10)])
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[]
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>>> max_tasks([])
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[]
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>>> max_tasks([(0, 10), (0, 20), (0, 30), (0, 40)])
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[]
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>>> max_tasks([(-1, 10), (-2, 20), (-3, 30), (-4, 40)])
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[]
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"""
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tasks = sorted(
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(
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Task(task_id, deadline, reward)
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for task_id, (deadline, reward) in enumerate(tasks_info)
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),
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key=attrgetter("reward"),
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reverse=True,
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)
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return [task.task_id for i, task in enumerate(tasks, start=1) if task.deadline >= i]
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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print(f"{max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)]) = }")
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print(f"{max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)]) = }")
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