Completed 1 problem

[lakshmidurgat]

No description provided.

[super30admin]

Interview Problem: Two Sum (Problem1.java)

Great work! Your solution is excellent and actually better than the reference solution in terms of time complexity. Here are some observations:

Strengths:

• You correctly implemented the HashMap-based approach, which is the optimal solution for this problem
• Your solution directly addresses the follow-up question asking for less than O(n²) time complexity
• The code is clean, readable, and well-structured
• You correctly handle edge cases like duplicate values (e.g., [3,3] with target 6)

Minor suggestions for improvement:

• Consider using the Map<Integer, Integer> interface type instead of HashMap<Integer, Integer> directly. This follows the "program to an interface" principle and makes the code more flexible.
• You could add a brief comment explaining the approach for future readers.
• Make sure to import java.util.HashMap (or java.util.Map) at the top of your file.

Overall: This is a high-quality solution that demonstrates strong understanding of the problem and efficient algorithm design. The trade-off between time and space complexity is well-handled.

VERDICT: PASS

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Interview Problem: 0-1 Knapsack Problem

It appears that the submitted solution addresses a completely different problem (Two Sum) rather than the 0-1 Knapsack problem described in the assignment. The 0-1 Knapsack problem requires:

1. Taking two arrays: weights[] and values[] (or profit[])
2. A knapsack capacity W
3. Finding the maximum total value of items that can fit within capacity W
4. Each item can either be included entirely or excluded (no partial inclusion)

The typical approach uses dynamic programming with a 2D table where:

• dp[i][j] represents the maximum value achievable using the first i items with capacity j
• For each item, you decide whether to include it or not based on whether its weight fits in the remaining capacity

Please review the problem statement carefully and implement a solution that:

• Iterates through items and capacities
• Uses the recurrence relation: dp[i][j] = max(dp[i-1][j], value[i-1] + dp[i-1][j-weight[i-1]])
• Returns the maximum value achievable

VERDICT: NEEDS_IMPROVEMENT