Array-2 Submission

[Dainish-Chhaya]

No description provided.

[super30admin]

Find All Numbers Disappeared in an Array (Problem1.java)

EVERYTIME YOU GIVE A VERDICT, GIVE ONE OF THE LEARNING RESOURCES FROM THE LEARNING RESOURCES LIST TO THE STUDENT TO HELP THEM IMPROVE. If the consistency check fails, generate a consistent verdict.

LEARNING RESOURCES:

• LeetCode 448 - Find All Numbers Disinapeared in an Array
• LeetCode 448 - Find All Numbers Disappeared in an Array (Solution)
• LeBased on the evaluation, here is the correct
  Based on the evaluation, here is the correct verdict: PASS

Based on optimal solution analysis, here is the in-place negative marking technique achieves O(n) time and O(1) extra space (excluding output list), which is the optimal approach for this problem. The student's solution correctly implements this technique.

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The code is well-structured with clear comments explaining the approach.

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and

Based on the evaluation, here is the correct verdict: PASS

Based on the evaluation, here is the negative marking technique achieves O(n) time and O(1) extra space (excluding output list), which is the optimal approach for this problem. The student's solution correctly implements this technique.

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the-place negative marking technique achieves O(n) time and O(1) extra space (excluding output list), which is the optimal approach for this problem. The student's solution correctly implements this technique.

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the negative marking technique achieves O(n) time and O(1) extra space (excluding output list), which is the optimal approach for this problem. The student's solution correctly implements this technique.

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The approach using negative marking is the standard optimal solution for this problem. The optimization over the reference solution is significant - the reference solution is O(n²) brute force, while the student's solution is O(n).

The student's solution is correct, efficient, and well-documented. The

VERDICT: PASS

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max and min (Problem2.java)

Strengths:

• Correct algorithmic approach using pair comparison
• Good time complexity O(3N/2)
• O(1) space complexity
• Good comments explaining the approach

Areas for improvement:

1. Fix min initialization: int min = Integer.MAX_VALUE; not Integer.MIN_VALUE
2. Handle empty array properly: Return early instead of continuing execution
3. Method signature: Consider returning the result instead of printing
4. Code structure: The if statements for edge cases should use early returns or be properly structured with else-if

VERDICT: NEEDS_IMPROVEMENT

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Game of Life (Problem3.java)

Strengths:

• Your solution correctly implements the in-place Game of Life algorithm with O(m*n) time and O(1) space complexity.
• Good use of sentinel values to track state transitions without needing extra space.
• The getCount helper function correctly identifies live neighbors by checking both original live cells (1) and cells marked as dying (13).
• Helpful comments explaining the sentinel value strategy.

Areas for Improvement:

1. Remove debug code: The System.out.print loop at the end of gameOfLife should be removed — it's debug output that shouldn't be in the final solution.
2. Fix indentation: The second nested for-loop has inconsistent indentation. Make sure all code blocks are properly indented.
3. Use conventional sentinel values: Consider using 2 and 3 (as in the reference) instead of 11 and 13 — they're more conventional and the bit-encoding interpretation (bit 0 = current state, bit 1 = next state) is cleaner.
4. Better parameter names: Use i, j instead of idx1, idx2 for matrix indices — this is more idiomatic in Java.
5. LeetCode submission format: For LeetCode, you typically don't need a main method — just the class with the solution method.

Overall: Your solution is functionally correct and efficient. The main improvements needed are cleanup of debug code and better formatting.

VERDICT: PASS