[Add] : Job Sequence program under GREEDY methods (#10482)

* to add job seq program

* to add job seq program

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* to add definitions in parameters

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* to add definitions in parameters

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* to add definitions in parameters

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* changes as recommended

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* type hint error resolved

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* removed lambda

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* 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>

scheduling/job_sequence_with_deadline.py

@@ -0,0 +1,62 @@
+"""
+Given a list of tasks, each with a deadline and reward, calculate which tasks can be
+completed to yield the maximum reward.  Each task takes one unit of time to complete,
+and we can only work on one task at a time.  Once a task has passed its deadline, it
+can no longer be scheduled.
+
+Example :
+tasks_info = [(4, 20), (1, 10), (1, 40), (1, 30)]
+max_tasks will return (2, [2, 0]) -
+Scheduling these tasks would result in a reward of 40 + 20
+
+This problem can be solved using the concept of "GREEDY ALGORITHM".
+Time Complexity - O(n log n)
+https://medium.com/@nihardudhat2000/job-sequencing-with-deadline-17ddbb5890b5
+"""
+from dataclasses import dataclass
+from operator import attrgetter
+
+
+@dataclass
+class Task:
+    task_id: int
+    deadline: int
+    reward: int
+
+
+def max_tasks(tasks_info: list[tuple[int, int]]) -> list[int]:
+    """
+    Create a list of Task objects that are sorted so the highest rewards come first.
+    Return a list of those task ids that can be completed before i becomes too high.
+    >>> max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)])
+    [2, 0]
+    >>> max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)])
+    [3, 2]
+    >>> max_tasks([(9, 10)])
+    [0]
+    >>> max_tasks([(-9, 10)])
+    []
+    >>> max_tasks([])
+    []
+    >>> max_tasks([(0, 10), (0, 20), (0, 30), (0, 40)])
+    []
+    >>> max_tasks([(-1, 10), (-2, 20), (-3, 30), (-4, 40)])
+    []
+    """
+    tasks = sorted(
+        (
+            Task(task_id, deadline, reward)
+            for task_id, (deadline, reward) in enumerate(tasks_info)
+        ),
+        key=attrgetter("reward"),
+        reverse=True,
+    )
+    return [task.task_id for i, task in enumerate(tasks, start=1) if task.deadline >= i]
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()
+    print(f"{max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)]) = }")
+    print(f"{max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)]) = }")