design-2

[spencerkrebs]

No description provided.

[super30admin]

Create Queue using Stacks (queue-using-stacks.py)

Strengths:

• The solution correctly uses two stacks to simulate a queue, with inStack for push operations and outStack for pop and peek operations.
• The time complexity for push is O(1), and for pop and peek, it is amortized O(1) because each element is moved from inStack to outStack at most once.
• The space complexity is O(n), which is optimal for this problem.
• The code is clean and well-commented, making it easy to understand.

Areas for Improvement:

• The empty method can be simplified. Instead of checking the lengths of both stacks and returning a boolean based on that, you can directly return the result of the condition not self.inStack and not self.outStack. This is more concise and efficient.
• While the comments are helpful, they are not strictly necessary since the code is self-explanatory. However, they are a good practice for clarity.

Suggested Improvement for empty:

def empty(self) -> bool:
    return not self.inStack and not self.outStack

This change makes the method more efficient and aligns with the reference solution.

Overall, the solution is excellent and only requires a minor improvement in the empty method for conciseness.

VERDICT: PASS

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Implement Hash Map (design-hashmap.py)

Your solution has a good start with the use of chaining and dummy nodes for each bucket. However, there are some issues that need to be addressed:

• In the get method, you are starting from the dummy node (which has key=-1) and checking the current node's key. Since the dummy node is always present and its key is -1, you will never match the target key. Instead, you should traverse the list by checking the next node's key, similar to how you do in the put and remove methods. For example, in get, you should do:

  cur = self.map[self.hash(key)]
  while cur.next:
      if cur.next.key == key:
          return cur.next.val
      cur = cur.next
  return -1

• Similarly, in the remove method, you are correctly checking the next node's key, but you are not handling the case where the key might be the first node after the dummy. Your current remove method should work because you are checking cur.next.key, but note that you are starting from the dummy node. However, you have an extra condition while cur and cur.next which is correct.

• In the put method, you are correctly traversing the list to update an existing key or append a new node. However, you should note that the dummy node is always present, so you don't need to check for the head being null.

• You initialized the map with 1000 dummy nodes. This is acceptable, but you should consider that the problem constraints allow up to 10^4 operations, so 1000 buckets might be too low. The reference solution uses 1000 buckets as well, so it's acceptable.

• Your hash function uses key % len(self.map), which is correct. However, note that len(self.map) is 1000, so the modulo operation will give a value between 0 and 999.

• It's good practice to use a dummy node to simplify the linked list operations, but you need to ensure all methods are consistent in their traversal.

To fix your solution:

1. Correct the get method to traverse by checking the next node.
2. Ensure the remove method correctly removes the node by adjusting pointers.
3. Test your solution with the provided example to ensure it works.

Here is a corrected version of your code:

class Node:
    def __init__(self, key=-1, val=-1, next=None):
        self.key = key
        self.val = val
        self.next = next

class MyHashMap:
    def __init__(self):
        self.map = [Node() for _ in range(1000)]
    
    def hash(self, key):
        return key % len(self.map)
    
    def put(self, key, value):
        index = self.hash(key)
        cur = self.map[index]
        while cur.next:
            if cur.next.key == key:
                cur.next.val = value
                return
            cur = cur.next
        cur.next = Node(key, value)
    
    def get(self, key):
        index = self.hash(key)
        cur = self.map[index]
        while cur.next:
            if cur.next.key == key:
                return cur.next.val
            cur = cur.next
        return -1
    
    def remove(self, key):
        index = self.hash(key)
        cur = self.map[index]
        while cur.next:
            if cur.next.key == key:
                cur.next = cur.next.next
                return
            cur = cur.next

VERDICT: NEEDS_IMPROVEMENT