cycle

[spencerkrebs]

No description provided.

[super30admin]

Reverse a LinkedList (reverse-linked-list.py)

Your solutions are correct and efficient. The iterative solution is optimal in both time and space. For the recursive solution, note that the space complexity is O(n) due to the recursion stack, not O(1). This is an important distinction, especially for large lists, as it could lead to stack overflow for very long lists.

You might consider adding comments to explain the recursive steps more clearly. For example, in the recursive solution, after the base case, the code sets head.next.next = head to reverse the link between the current node and the next node. Then it sets head.next = None to break the old link. This is standard, but a brief comment could help readability.

Also, ensure that your complexity analysis comments are accurate. The recursive solution is O(n) time and O(n) space.

Overall, great job providing both iterative and recursive solutions as suggested in the follow-up.

VERDICT: PASS

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Remove nth node (nth-node.py)

Strengths:

• The solution correctly implements the two-pointer technique to remove the nth node from the end in one pass, which is efficient and meets the follow-up requirement.
• The use of a dummy node simplifies handling edge cases, like when the head needs to be removed.
• The code is concise and readable, with clear variable names (dummy, right, left) and logical steps.

Areas for Improvement:

• While the code is correct, it could benefit from a brief comment explaining the purpose of the dummy node and the two-pointer approach, especially for clarity in an interview setting.
• The variable names left and right are common for two-pointer techniques, but consider using more descriptive names like fast and slow to indicate their roles (though right is often the fast pointer and left the slow one in this context).
• The loop for i in range(n) moves the right pointer n steps ahead, which is correct. However, note that the dummy node is used, so the initial step is from dummy. This ensures that when right reaches the end, left is exactly at the node before the one to be removed.
• The while loop condition while right.next is correct, but it's important to note that right starts at dummy and moves n steps, so it may be at the last node or beyond? Actually, since n is at least 1 and at most the size of the list, and dummy is added, the initial move of n steps from dummy is safe. However, if n equals the size of the list, then after moving n steps, right will be at the node before the last? Wait, let's clarify: the list has size L. The dummy node is added, so the total nodes become L+1. Moving n steps from dummy, right will be at the n-th node from the start (0-indexed). Then, when we move both until right.next is None, left will be at the (L-n)-th node from the start (which is the node before the one to remove). This is correct.

But note: if n is equal to the size of the list (L), then after moving n steps, right will be at the L-th node (which is the last node) because dummy is node0, then node1, ... nodeL. Then right.next is None, so the while loop doesn't run. Then left is still at dummy. Then we remove dummy.next, which is head. So it correctly removes the head.

So the code is correct.

One minor point: in Python, it's common to use _ for unused loop variables. So the loop could be written as for _ in range(n): to indicate that the index is not used.

Overall, the solution is excellent and optimal.

VERDICT: PASS

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Cycle in linked list (linked-list-cycle-ii.py)

Your solution is correct and efficient. You've successfully implemented Floyd's algorithm, which is the standard approach for this problem. Here are some strengths and minor suggestions:

Strengths:

• You correctly handle the edge case where the list is empty.
• You use two pointers (slow and fast) to detect a cycle, which is optimal.
• After detecting a cycle, you reset one pointer to the head and move both at the same pace to find the cycle start.
• Your code is clean and well-commented, making it easy to understand.

Areas for improvement:

1. The variable name found could be more descriptive. Consider renaming it to hasCycle or cycleDetected to make its purpose clearer.
2. In Python, it's common to use is for comparing with None, but for node comparisons, == is appropriate. However, note that in Python, is checks for identity (same object), while == checks for equality (same value). Since nodes are typically compared by identity in linked list problems (as we care about the same node object, not just same value), using is might be more precise. For example:
   • Change if fast == slow to if fast is slow
   • Similarly, change while pointer != slow to while pointer is not slow
     This avoids potential issues if nodes have the same value but are different objects.
3. You can avoid the found flag by checking the condition after the loop. For example:

   if fast is None or fast.next is None:
        return None

   Then you don't need the flag. However, your current approach with the flag is also acceptable.

Overall, your solution is excellent and follows best practices. The suggestions are minor and primarily stylistic.

VERDICT: PASS