S30 FAANMG Problem #4 | Create Queue using Stacks

[sandy7907]

No description provided.

[super30admin]

• The solution is correct and efficient. It correctly implements a queue using two stacks with amortized O(1) time for operations.

• The code is readable and well-structured.

• However, there is a syntax error: the class is missing a closing brace. This would cause a compilation error. Please add a closing brace at the end of the file.

• Additionally, to avoid code duplication, you can consider having the pop method call the peek method to ensure the outStack is updated, and then pop. For example:

  public int pop() {
      peek(); // This ensures outStack has the front element if available
      return outStack.pop();
  }
  

  This way, the transfer logic is only in peek. But the current approach is also acceptable.

Overall, the solution is good except for the missing brace.

[super30admin]

Strengths:

• The solution correctly implements the queue using two stacks.
• The logic for transferring elements from inStack to outStack when outStack is empty is correctly implemented in both pop and peek.
• The code is clean and easy to understand.

Areas for improvement:

• Although the problem states that all calls to pop and peek are valid, it is a good practice to consider edge cases. For example, if pop or peek is called when both stacks are empty, the code will throw an exception (like EmptyStackException). You might want to handle this by returning a specific value or throwing a more descriptive exception, but given the problem constraints, it is not strictly necessary.
• The pop method in the reference solution first calls peek to ensure the outStack has elements, which avoids code duplication. In your solution, both pop and peek have the same code for transferring elements. You could refactor to avoid duplication by having pop call peek first, like the reference solution does. This would make the code more maintainable.

Suggested improvement for pop:

public int pop() {
    peek(); // Ensure outStack has the front element
    return outStack.pop();
}

This way, the transfer logic is only in peek, and pop reuses it.