Design-1 Solution

[indrasena483]

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[super30admin]

Implement Hash Set (HashSet.java)

END OF FORMAT

Remember: The student's solution is in Java, and the reference solution is also in Java. The problem is to implement a HashSet without using built-in hash table libraries.

Let's evaluate the student's solution step by step.

First, the student's code:

class MyHashSet {
    int primaryBuckets;
    int secondaryBuckets;
    boolean[][] storage;

    public MyHashSet() {
        primaryBuckets = 1000;
        secondaryBuckets = 1000;
        storage = new boolean[primaryBuckets][];
    }

    private int getPrimaryHash(int key) {
        return key % primaryBuckets;
    }

    private int getSecondaryHash(int key) {
        return key / secondaryBuckets;
    }

    public void add(int key) {
        int primaryIndex = getPrimaryHash(key);
        if (storage[primaryIndex] == null) {
            if (primaryIndex == 0) {
                storage[primaryIndex] = new boolean[secondaryBuckets + 1];
            } else {
                storage[primaryIndex] = new boolean[secondaryBuckets];
            }
        }
        int secondaryIndex = getSecondaryHash(key);
        storage[primaryIndex][secondaryIndex] = true;
    }

    public void remove(int key) {
        int primaryIndex = getPrimaryHash(key);
        if (storage[primaryIndex] == null) {
            return;
        }
        int secondaryIndex = getSecondaryHash(key);
        storage[primaryIndex][secondaryIndex] = false;
    }

    public boolean contains(int key) {
        int primaryIndex = getPrimaryHash(key);
        if (storage[primaryIndex] == null) {
            return false;
        }
        int secondaryIndex = getSecondaryHash(key);
        return storage[primaryIndex][secondaryIndex];
    }
}

Now, let's evaluate based on the criteria:

1. Correctness:

   • The student is using a 2D boolean array to represent the HashSet. The idea is to use double hashing to map keys to indices.
   • The primary hash function is key % primaryBuckets (which is 1000), and the secondary hash function is key / secondaryBuckets (which is also 1000).
   • However, there is a potential issue: when primaryIndex == 0, the student is creating an array of size secondaryBuckets + 1 (i.e., 1001), and for other indices, they are creating an array of size secondaryBuckets (i.e., 1000). This seems arbitrary and might lead to inconsistencies.
   • Why is there a special case for index 0? The student might be trying to handle the case when key is 0? But note that for key=0, getSecondaryHash(0) returns 0, which is within the range [0, 1000) if the array size is 1000? But wait, if the array for index0 is of size 1001, then index 0 to 1000 are valid. For other indices, the array is of size 1000, so indices 0 to 999 are valid.
   • However, the secondary hash function key / 1000 can produce values from 0 to 1000000/1000 = 1000. So for keys from 0 to 999999, getSecondaryHash(key) returns a value from 0 to 999. But for key=1000000, it returns 1000. So the secondary index can be 1000 only for keys >= 1000000.
   • But the problem states that key is between 0 and 10^6. So the maximum key is 1000000. For key=1000000, getSecondaryHash(1000000) = 1000000/1000 = 1000. So we need an array that can index up to 1000. Therefore, for each primary bucket, we need a secondary array of size 1001 (to have indices 0 to 1000).
   • The student has correctly identified this for index0: they create an array of size 1001. But for other indices, they create an array of size 1000. This is incorrect because for any primary index, we might have keys that require secondary index 1000. For example, key=1000:
     primaryIndex = 1000 % 1000 = 0 -> secondaryIndex = 1000/1000 = 1 -> which is within [0,1000] for the array of size 1001? Wait, for key=1000:
     primaryIndex = 1000 % 1000 = 0 -> secondaryIndex = 1000/1000 = 1 -> which is valid for an array of size 1001 (indices 0 to 1000).
   • But key=2000: primaryIndex=0, secondaryIndex=2 -> also valid.
   • Now consider key=1000000: primaryIndex=0, secondaryIndex=1000 -> which is valid for the array of size 1001.
   • However, what about keys that map to primaryIndex=1? For example, key=1001:
     primaryIndex = 1001 % 1000 = 1
     secondaryIndex = 1001/1000 = 1 -> which is within [0,999] for an array of size 1000? But wait, the array for primaryIndex=1 is of size 1000, so indices 0 to 999. So secondaryIndex=1 is valid.
   • But key=2001: primaryIndex=1, secondaryIndex=2 -> valid.
   • The problem arises for keys that are multiples of 1000 but not zero? For example, key=1000: maps to primaryIndex=0 -> which has an array of size 1001 -> secondaryIndex=1 -> valid.
   • However, consider key=1000000: which maps to primaryIndex=0 -> secondaryIndex=1000 -> valid.
   • But what about key=0?
     primaryIndex=0, secondaryIndex=0 -> valid.
   • So why does the student have a special case for primaryIndex=0? Actually, for every primary index, we need a secondary array of size 1001 because the secondary index can be from 0 to 1000 (since key max=1000000, and 1000000/1000=1000). So for any primary index, the secondary array should be of size 1001.
   • The student's code only allocates size 1001 for primaryIndex=0, and for others it allocates 1000. This means that for keys with primaryIndex not 0, if the secondaryIndex is 1000, it will cause an array index out of bounds.
   • For example, key=1000000: primaryIndex=0 -> okay.
   • But key=999000:
     primaryIndex = 999000 % 1000 = 0? Wait, 999000 % 1000 = 0 -> so it maps to index0 -> which has size 1001 -> secondaryIndex=999000/1000=999 -> valid.
   • Actually, keys that would have secondaryIndex=1000 must be >=1000000, but the maximum key is 1000000. So only key=1000000 has secondaryIndex=1000. And key=1000000 has primaryIndex=0. So for primaryIndex=0, we have allocated size 1001, which covers secondaryIndex=1000. For other primary indices, the maximum secondaryIndex is 999 (because for key=999999: primaryIndex=999999%1000=999, secondaryIndex=999999/1000=999). So for primaryIndex=1 to 999, we only need an array of size 1000 (indices 0 to 999).
   • Therefore, the student's approach is correct: for primaryIndex=0, we need an array of size 1001 to accommodate secondaryIndex=0 to 1000. For other primary indices, we only need an array of size 1000 because the secondaryIndex will be at most 999.
   • So the code should work correctly for all keys in [0,1000000].

2. Time Complexity:

   • Each operation (add, remove, contains) does constant time: two hash calculations and array accesses. So O(1) per operation. This matches the reference solution.

3. Space Complexity:

   • The storage is a 2D array with 1000 primary buckets. For index0, we have 1001 elements, and for indices 1 to 999, we have 1000 elements each. So total space = 1001 + 999*1000 = 1001 + 999000 = 1000001. So O(10^6) which is acceptable given the constraints. The reference solution also uses O(10^6) space.

4. Code Quality:

   • The code is well-structured and readable.
   • There are a few typos: primaryBuckets is misspelled as primaryBuckets in the class variable declaration, but correctly spelled in the constructor? Actually, in the code:
     int primaryBuckets; -> correct spelling?
     But in the constructor: primaryBuckets = 1000; -> wait, the variable is declared as primaryBuckets? Actually, in the student's code:
     int primaryBuckets; -> but then in the constructor: `primaryBuckets =

VERDICT: NEEDS_IMPROVEMENT

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Implement Min Stack (MinStack.java)

Your solution is well-structured and meets the problem requirements. Here are some points to consider:

1. Initialization: You initialized two stacks without any initial values. The reference solution initializes minStack with Integer.MAX_VALUE to ensure that the first push operation sets the min correctly. While your code handles the first push by checking if minStack is empty, initializing minStack with a large value can make the code more consistent and avoid potential issues if getMin were called on an empty stack (though the problem states it won't happen). However, your current approach is correct given the constraints.

2. Efficiency: Your solution is efficient with O(1) time for all operations and O(n) space. The minStack stores the same minimum value repeatedly when consecutive pushes have the same minimum, which is acceptable and does not affect the time complexity.

3. Code Quality: The code is clean, readable, and well-commented. You explained the approach clearly. However, you can consider adding a comment in the constructor to explain why minStack is not initialized with a value, or you could initialize it for clarity.

4. Edge Cases: Your code handles the push and pop operations correctly. Since the problem states that pop, top, and getMin are called on non-empty stacks, you don't need to worry about empty stacks in those methods. But for real-world scenarios, you might want to add error handling (like throwing exceptions) for empty stacks, though it's not required here.

Overall, great job! Your solution is correct and efficient.

VERDICT: PASS