Add implementation for PreCourse-1

[rishigoswamy]

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[super30admin]

Let's evaluate each exercise one by one.

Exercise_1.py (Array-based Stack Implementation):

• Correctness: The implementation correctly simulates a stack using a list. The methods (isEmpty, push, pop, peek, size, show) behave as expected for a stack. The solution handles stack overflow and underflow with appropriate messages.
• Time Complexity: The student correctly identified the time complexities: O(1) for all operations except show which is O(n). This is accurate.
• Space Complexity: The student correctly states O(n) for space, which is correct.
• Code Quality: The code is well-structured and readable. However, there are a few minor issues:
  • The isEmpty method uses return not self.stack, which is correct, but note that self.currentSize is also maintained. It might be more consistent to use return self.currentSize == 0.
  • In the pop and peek methods, when the stack is empty, they print "Stack Underflow" and return 0. However, returning 0 might be problematic if the stack is intended to hold any data type (e.g., if it holds strings, returning 0 is inconsistent). It would be better to return None or raise an exception.
  • The show method prints each item and then returns the list. This might be confusing because the problem says "show" should probably just display the stack. Returning the list could be useful, but the problem description isn't clear. However, it's not a major issue.
• Efficiency: The implementation is efficient. The list is used as a dynamic array, and appending and popping from the end are O(1) operations.

Exercise_2.py (Linked List-based Stack Implementation):

• Correctness: The implementation uses a linked list with a dummy head node. The push and pop operations are correctly implemented at the head. The interactive part works as expected.
• Time Complexity: The student correctly identifies O(1) for push and pop.
• Space Complexity: O(n) is correct.
• Code Quality: The code is readable. However:
  • The dummy node is initialized with self.head = Node(None). This is acceptable, but note that the dummy node itself doesn't hold data. The actual top of the stack is self.head.next. This is a common technique, but it might be slightly overkill for a stack. Alternatively, we could have self.head directly point to the top node (without a dummy). However, the current implementation is correct.
  • In the pop method, if the stack is empty, it returns None. The interactive part handles None correctly by printing "Stack is empty." This is good.
  • The push method creates a new node and correctly updates pointers.
• Efficiency: The implementation is efficient with O(1) for both operations.

Exercise_3.py (Singly Linked List Implementation):

• Correctness:
  • The append method adds at the end. It correctly handles the empty list case and traverses to the end. This is correct.
  • The find method searches linearly and returns the node or None. This is correct.
  • The remove method uses a dummy node to handle the case of removing the head. It correctly removes the first occurrence. However, note that after removal, it updates self.head = dummy.next only once at the end. This is correct because if we remove the head, the dummy's next becomes the new head. But note: the code sets self.head = dummy.next every time, even when the head didn't change. This is redundant but not wrong. Alternatively, we could update self.head only when removing the head? Actually, the dummy node approach is elegant because it avoids special cases for the head. However, the code always sets self.head = dummy.next after every removal. This is unnecessary if the head didn't change, but it doesn't break anything. It would be more efficient to only update if we removed the head, but the current code is correct.
• Time Complexity: The student correctly identifies O(n) for append, find, and remove.
• Space Complexity: O(n) is correct.
• Code Quality: The code is readable. However:
  • In the remove method, the line self.head = dummy.next is executed in every successful removal. This is acceptable but note that if the list is long and we remove an element that is not the head, we are updating self.head to the same value it already had. It would be better to only update if we are removing the head. Alternatively, we can avoid the dummy node and handle the head removal separately. But the current approach is correct.
  • The remove method returns without any value, which is standard for such operations.
• Efficiency: The operations are as efficient as possible for a singly linked list without tail pointer. For append, without a tail pointer, it is O(n). If we were to optimize appends, we could maintain a tail pointer, but the problem doesn't require it.

Overall, all three solutions are correct and efficient. The code quality is good with minor issues that don't affect correctness.