Competitive Coding 2

[MeghaN28]

No description provided.

[super30admin]

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

First, I need to evaluate the student's solution for the two-sum problem. The problem is to find two indices in an array where the elements at those indices add up to a target. The student's solution uses a HashMap to store numbers and their indices, and for each element, it checks if the complement (target - current number) exists in the map.

Let's look at the code:

```java
class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> hashnum = new HashMap<>();

        for (int i = 0; i < nums.length; i++) {
            int num = target - nums[i];

            if (hashnum.containsKey(num)) {
                return new int[] { hashnum.get(num), i };
            } else {
                if (!hashnum.containsKey(nums[i])) {
                    hashnum.put(nums[i], i);
                }
            }
        }
        return null;
    }
}

Now, I'll evaluate based on the criteria:

1. Correctness: Does the solution solve the problem correctly?

   • The idea is correct: use a map to store each number and its index. For each number, calculate the complement and check if it's in the map. If yes, return the indices.
   • However, there's a potential issue: the else block only adds the current number to the map if it's not already present. This might be problematic if there are duplicate numbers. For example, if the array has duplicates like [3, 3] and target is 6, it should return [0,1]. But in this code, when i=0, nums[i]=3, it's not in the map, so it adds (3,0). Then when i=1, nums[i]=3, complement is 3, and it should be in the map. So it should work. But let's see: for i=1, it checks if hashnum contains key (target - nums[1]) = 6-3=3. Yes, it does, with value 0. So it returns [0,1]. Good.
   • Another case: [1,2,2,3], target=4. It should find 2 and 2? But wait, the problem says "you may not use the same element twice." So if there are two 2s, indices 1 and 2, it should be able to use them since they are different elements. In this code, when i=1, nums[1]=2, complement=2, but the map doesn't have 2 yet? Let's simulate:
     i=0: num=1, complement=3? target=4, nums[0]=1, so complement=3. Not in map. Then check if nums[0]=1 is in map? No, so add (1,0).
     i=1: nums[1]=2, complement=2. Check if map has key 2? No, because we haven't added it yet. Then in else block, check if nums[1]=2 is in map? No, so add (2,1).
     i=2: nums[2]=2, complement=2. Now map has key 2? Yes, with value 1. So it returns [1,2]. Correct.
   • But what if the duplicate is not needed? For example, [2,2,3], target=4. It should return [0,1]. Same as above, it will work.
   • However, there's a subtle issue: the code only adds the current number to the map if it's not already present. This might be intentional to avoid overwriting, but actually, for two-sum, if there are duplicates, we might miss a solution if we don't update the index? No, because we are iterating from left to right, and when we find a duplicate, we don't update the map. But that's okay because we only need one solution, and the first occurrence is stored. When we check for complement, we are looking for a number that we have seen before, which is stored with its index. So if we have duplicates, we don't need to update the map because the first index is sufficient to form a pair with a later duplicate. For example, in [3,3], target=6: first 3 is stored at index 0. When we see second 3 at index 1, we check complement=3, which is in map with index 0, so we return [0,1]. If we had [3,4,3], target=6: first 3 at index0 stored. Then at index1, 4, complement=2, not in map, then add 4 at index1. Then at index2, 3, complement=3, which is in map with index0, so return [0,2]. Correct.
   • But what if the target is 7 and array is [3,4,3]? Then at index0: complement=4, not in map, add (3,0). Index1: nums[1]=4, complement=3, which is in map, so return [0,1]. Correct.
   • So the solution seems correct. However, there is a return null at the end, which is fine since the problem states there is exactly one solution.

2. Time Complexity: How does it compare to the reference solution?

   • The reference solution is O(n^2) with nested loops.
   • This solution uses a single loop and HashMap operations which are O(1) on average. So overall time complexity is O(n), which is better than O(n^2). So it's an improvement.

3. Space Complexity: How does it compare to the reference solution?

   • Reference solution uses O(1) space.
   • This solution uses a HashMap that can store up to n elements in the worst case, so O(n) space. So it uses more space, but trades off for better time.

4. Code Quality: Is the code well-structured, readable, and following best practices?

   • The code is mostly readable. Variable names could be better: hashnum might be better as numToIndex or something more descriptive.

   • The logic is a bit nested. The else block contains an if statement, which could be simplified.

   • Actually, the else block is unnecessary. We can always put the current number into the map after checking the complement, but only if we don't have it already? Actually, no: we should not add the current number before checking the complement, because if we add it first, then when the complement is the same as current, it might find itself. For example, if target is 6 and current number is 3, if we add it to the map and then check complement=3, it would find itself, which is not allowed since we can't use the same element twice. So we must check first and then add.

   • In this code, the adding is done only if the complement is not found and only if the number is not already in the map. But actually, we can add the number regardless of whether it's already in the map or not, but we don't need to because if it's already in the map, we don't want to overwrite the index? Actually, we don't need to overwrite because the first index is sufficient. So the code is correct in that regard.

   • However, the code can be written more clearly without the nested if in else. For example:
     if (map.containsKey(complement)) {
     return new int[]{map.get(complement), i};
     }
     map.put(nums[i], i);

   • This is the standard way. The student has an extra check: they only put if the number is not already in the map. But as we saw, it's not necessary to check because if the number is already in the map, we don't need to update it. So we can simply put the current number after the check. This would be simpler.

   • Also, in the student's code, if the complement is not found, they check if the current number is not in the map before adding. But if the number is already in the map, we skip adding. This is correct because we want to keep the first occurrence. But actually, if we have duplicates, we don't need to add the duplicate because the first index is already there. So it's efficient to avoid duplicate keys, but since HashMap doesn't allow duplicate keys anyway, if we do map.put(nums[i], i), it will overwrite the previous index. That would be bad because we might lose the first index. For example, if we have [3,3] and we put index1 for key3, then if we have a later number that needs 3, it would find index1, but if we had stored index0, it would be better? Actually, no: if we overwrite, then for a duplicate, we store the later index, but when we check for complement, we might need the earlier index. So we should not overwrite. Therefore, the student's approach of only adding if not present is correct and necessary.

   • However, the standard solution usually does not check for existing key when putting; instead, it avoids overwriting by not putting again. But in code, we can do:
     if (!map.containsKey(nums[i])) {
     map.put(nums[i], i);
     }

   • Or we can iterate and only add after checking complement, which is what the student did.

   • But in the student's code, the structure is:
     if (hashnum.containsKey(num)) {
     return ...;
     } else {
     if

VERDICT: NEEDS_IMPROVEMENT

---

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

Strengths:

• The solution correctly implements the dynamic programming table and fills it in a bottom-up manner.
• The comments are helpful and explain the approach well.
• The code is readable and uses meaningful variable names.

Areas for Improvement:

• Consider renaming the method parameters to be more descriptive (e.g., capacity instead of W, values instead of val, weights instead of wt). This improves readability.
• The method name knapsack is quite generic. If this is part of a larger project, consider a more specific name like findMaxValue or similar.
• While not necessary, you could mention the possibility of space optimization using a 1D DP array for future reference.
• The code does not handle edge cases explicitly (e.g., if the input arrays are empty), but the current implementation handles them correctly because the DP table is initialized with zeros. However, adding a comment about this might be beneficial.

Overall, the solution is correct and efficient. Good job!

VERDICT: PASS