Completed Competitive-Coding-10

[sarvanibaru]

No description provided.

[super30admin]

Interview Problem : Buy and Sell Stock (BuyAndSellStock2.java)

Your solution is excellent! You have implemented an optimal greedy algorithm that efficiently computes the maximum profit by capturing all positive price differences. Here are some strengths and areas for improvement:

Strengths:

• Correctness: Your solution correctly handles all cases, including the examples provided.
• Efficiency: The time complexity is O(n) and space complexity is O(1), which is the best possible for this problem.
• Code Quality: The code is clean, simple, and well-commented. The approach is clearly explained.

Areas for Improvement:

• While your solution is perfect for this problem, it's worth noting that the recursive reference solution is provided for educational purposes to show a naive approach. In practice, your solution is the way to go.
• There are no significant improvements needed. However, you might consider adding a brief comment explaining why the greedy approach works (i.e., because you can engage in multiple transactions and every positive increment is profit).

Overall, great job!

VERDICT: PASS

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Interview Problem: Peeking Iterator (PeekingIterator.java)

The student has done a great job. The code is correct, efficient, and well-commented. However, one minor point: in the next() method, the student wrote if(itr.hasNext()) without a space, which is a style issue but not critical. Also, the constructor might benefit from a check for empty iterator, but given constraints, it's not necessary.

Verdict should be PASS.

But let's confirm the problem statement: it says "All the calls to next and peek are valid." meaning we don't have to worry about calling peek or next when there's no element? Actually, the problem says "Each language may have a different implementation", but in Java, the Iterator interface might throw NoSuchElementException if next is called when hasNext is false. However, in this implementation, since we have nextEl initialized with the first element, and we only call next on the underlying iterator when we know it has next, it should be fine.

Also, the student's hasNext() returns true only if nextEl != null, which is correct.

So, overall, no issues.

Final evaluation:

EVALUATION:
The student's solution is correct and efficient. It correctly implements the peeking iterator by maintaining a nextEl field that always holds the next element to be returned. The peek() method returns this value without advancing, while next() returns it and then advances the underlying iterator. The time and space complexities are optimal at O(1) for all operations. The code is well-structured and readable, with appropriate comments explaining the approach.

FEEDBACK:
Excellent job! Your solution is identical to the reference implementation and handles all requirements correctly. The code is clean and efficient. One minor stylistic note: consider adding a space after if for better readability (e.g., if (itr.hasNext())), but this is a small detail. Also, while not required due to constraints, in a more general case, you might want to handle the possibility of an empty iterator in the constructor, but since the problem guarantees at least one element, it's acceptable as is. Keep up the good work!

VERDICT: PASS