[Add] : Job Sequence program under GREEDY methods

greedy_methods/job_sequence.py

@@ -0,0 +1,58 @@
+"""
+Given a list of tasks, each with a deadline and reward,
+The maximum number of tasks that can be scheduled to maximize reward.
+We can only complete one task at a time, and each task takes 1 unit
+of time to complete. 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]) -
+which is by scheduling the tasks with rewards 40, 20
+
+This problem can be solved using the concept of "GREEDY ALGORITHM".
+Time Complexity - O(n log n)
+
+We iterate over the tasks array once, sorting it in descending order of reward.
+Then we iterate over the sorted tasks array once more, scheduling each
+task if its deadline is greater than the current time.The greedy choice at
+each point is to either schedule the current task if its deadline is greater
+than the current time, or skip it otherwise.
+"""
+
+
+class Task:
+    def __init__(self, id : int, deadline : int, reward : int):
+        self.id = id
+        self.deadline = deadline
+        self.reward = reward
+
+
+def max_tasks(tasks_info: list[tuple[int]]) -> int:
+    """
+    >>> max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)])
+    (2, [2, 0])
+    >>> max_tasks([(1, 10), (2, 20), (3, 30), (2, 40)])
+    (2, [3, 2])
+    """
+    tasks = [Task(i, d, p) for i, (d, p) in enumerate(tasks_info)]
+
+    tasks.sort(key=lambda task: task.reward, reverse=True)
+
+    schedule = []
+    current_time = 0
+
+    for task in tasks:
+        if task.deadline > current_time:
+            schedule.append(task.id)
+            current_time += 1
+
+    return len(schedule), schedule
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()
+
+    print(max_tasks([(4, 20), (1, 10), (1, 40), (1, 30)]))