Interview Problem : Two elements sum to target (Problem1.java)

Your solution is on the right track and demonstrates a good understanding of using hash maps to reduce time complexity. Here are some points to consider:

• Correctness: The problem guarantees exactly one solution, so your return statement for the no-solution case might never be executed. However, to be safe and consistent with the return type (which expects an array of two integers), you should return new int[]{-1, -1} instead of new int[]{-1}. This ensures that the array has two elements, which is expected.

• Variable Naming: Consider using more descriptive variable names. For example, instead of val, use complement to clearly indicate that it represents the value needed to pair with nums[i] to reach the target.

• Code Style: In Java, it's common to place the opening brace { on the same line as the class or method declaration. Also, ensure consistent indentation (which is mostly good here).

• Efficiency: Your approach is efficient with O(n) time complexity. However, you can note that in the best case (if the solution is found early), it might not need to process the entire array. This is already handled by your early return.

• Edge Cases: Your solution handles duplicate values correctly because you check for the complement before adding the current element to the map. This prevents using the same element twice.

Overall, you have a good solution that meets the follow-up challenge of achieving less than O(n^2) time complexity. With minor adjustments, it will be robust and follow Java best practices.

VERDICT: PASS

Interview Problem: 0-1 Knapsack Problem (Problem2.java)

Your solution is correct and follows the standard dynamic programming approach for the 0-1 knapsack problem. Well done! Here are a few points to consider for improvement:

1. Initialization: Your DP table initialization is correct because Java initializes arrays to zero by default. However, explicitly initializing the first row and first column to zero (if needed) can make the code more explicit and self-documenting, though it is not necessary here.

2. Loop Index for Capacity: In the inner loop, you start j from 0. This is acceptable, but note that for j < weight[i-1], you are only doing the "not choose" case, which is already set by dp[i][j] = dp[i-1][j]. So it's efficient. However, you could start j from weight[i-1] to W when doing the "choose" part to avoid the check for if (weight[i-1] <= j), but that would require restructuring the code. The current way is standard and clear.

3. Space Optimization: For large W, the space usage might be high. You could consider optimizing to use a 1D array to reduce space complexity to O(W). This is an advanced optimization but worth knowing. For example:

   int[] dp = new int[W+1];
   for (int i = 0; i < n; i++) {
        for (int j = W; j >= weight[i]; j--) {
            dp[j] = Math.max(dp[j], val[i] + dp[j - weight[i]]);
        }
   }
   return dp[W];
   

   This uses a single array and iterates backwards to avoid overwriting values needed for the current item.

4. Edge Cases: Consider handling edge cases such as when W is 0, or when there are no items. Your code should work for these because when n=0, the loop won't run and return dp[0][W]=0, which is correct. Similarly, if W=0, the inner loop from j=0 to 0 will set all values to 0. So it's robust.

5. Variable Names: While your variable names are good, using W for capacity and weight for the array of weights is clear. However, in the problem, the parameters are often named wt for weights and val for values, but your choice is also acceptable.

Overall, your solution is excellent. Keep up the good work!

VERDICT: PASS