Desig-2 Completed

[CHINMAY-PRAJAPATI]

No description provided.

[super30admin]

Create Queue using Stacks (implement_queue_using_stack.py)

Your solution is excellent and correctly implements the queue using two stacks. Here are some minor points for improvement:

1. The docstring at the top of the file is good practice, but note that it is placed outside the class. However, the problem does not require docstrings, so this is optional. If you include docstrings, consider adding them for the class and methods as well for better documentation.
2. In the peek method, you check if len(self.outstack) == 0. This is correct, but you can also write it as if not self.outstack for better readability and Pythonic style. Similarly, while len(self.instack) != 0 can be while self.instack.
3. The pop method calls self.peek() which transfers elements if needed, but note that peek returns the front element without popping. Then pop pops from outstack. This is efficient and avoids code duplication. However, ensure that if the queue is empty, pop should handle it. Your pop method relies on peek to transfer elements, but if both stacks are empty, peek will try to return self.outstack[-1] which will cause an index error. The reference solution checks empty() in pop to avoid this. Your solution does not have this check, so if pop is called on an empty queue, it will crash. The problem states that all calls to pop and peek are valid, so this might not be an issue, but for robustness, you could add a check. Since the constraints say all calls are valid, it is acceptable.

Overall, your code is clean and efficient. Well done!

VERDICT: PASS

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Implement Hash Map (hashmap.py)

Strengths:

• The code is clean and easy to read.
• The use of a separate ListNode class is appropriate.
• The hash function is simple and effective.

Areas for Improvement:

1. In the put method, you start from the dummy node and then check current.next. This is good, but when updating an existing key, you only check from the next node. However, if the key is the first node in the list (after the dummy), it will not be updated. For example, if you have a bucket with a dummy node and one node with key=5, and you try to put key=5 again, your code will not find it because current.next is the node with key=5, but you are checking current.next.key only after moving to the next. Actually, let me reexamine:

   • In put, you start with current = self.map[self.hash(key)] (which is the dummy node). Then you iterate while current.next exists. For the first node, you check if current.next.key == key -> if true, update and return. So that should work. But wait: what if the list has only the dummy node? Then current.next is None, so you don't enter the loop. Then you add a new node. That is correct.
   • However, if the key exists in the first node (after dummy), you do check it in the first iteration. So it seems correct. But note: your code does not check the dummy node itself (which has key=-1) so that is fine.

   Actually, upon closer inspection, the put method is correct. It starts at the dummy node and checks the next node. So it will update if the key exists in the first real node.

2. The get method starts from self.map[self.hash(key)].next, which skips the dummy node. This is acceptable because the dummy node doesn't store a key. However, it is consistent with the put method which also ignores the dummy node.

3. The remove method has an issue: it starts at the dummy node (which is good) and checks current.next. But if the key to remove is the first node (after dummy), it will be correctly removed. However, the code might not handle the case when the key is not found? It will traverse until the end and do nothing, which is correct.

4. One potential issue: the initializer creates 1000 dummy nodes. This is acceptable. But note that the reference solution uses 1000 buckets and creates dummy nodes only when needed (lazy initialization). Your solution pre-allocates all 1000 dummy nodes, which uses more memory initially. However, since the problem allows up to 10^4 calls, 1000 dummy nodes is acceptable.

5. Another minor issue: in the put method, if the key does not exist, you traverse the entire list to the end before adding. This is efficient enough.

6. The remove method might have a problem: when you remove a node, you set current.next = current.next.next. This is correct. But you should also break out of the loop after removal? Actually, you return immediately after removal, which is good.

Overall, the solution is mostly correct. However, there is one critical issue: the get method might not work correctly if the bucket is empty. For example, if there are no nodes in a bucket (only the dummy node), then self.map[self.hash(key)].next is None, so the while loop doesn't run and returns -1. This is correct.

But wait: what if the key is stored in the dummy node? It never is because the dummy node has key=-1. So it's safe.

After careful review, the code appears to be correct. However, let's test with an example:

• Put (1,1): hash(1)=1. The bucket at index1 has a dummy node. Then we set current = dummy. current.next is None, so we don't enter the while loop. Then we add a new node after dummy.
• Get(1): hash(1)=1. Then current = dummy.next (which is the node with key=1). Then we return its value (1). Correct.
• Remove(1): hash(1)=1. current = dummy. current.next exists and key=1? Then we remove it. Correct.

So the code is correct.

But one more edge: putting the same key twice.

• Put(1,1): as above.
• Put(1,2): hash(1)=1. current = dummy. current.next exists (key=1) -> so we update the value to 2. Correct.

Therefore, the solution is correct.

VERDICT: PASS