Complete Design-2 assignment - #2494

Implement Queue using Stacks (ImplementQueueUsingStacks.java)

Strengths:

• Correct implementation of the two-stack approach
• Good time and space complexity analysis in comments
• Clean, readable code structure
• Proper use of Java's Stack class
• Correct handling of edge cases (empty queue check)

Areas for improvement:

• Could add more inline comments explaining the logic
• The peek() method could be slightly more efficient by checking if secondStack is empty before the while loop
• Consider using Deque instead of Stack (Stack is considered legacy in Java)

VERDICT: PASS

Design HashMap (ImplementHashMap.java)

E attempted to use double hashing with a 2D array approach, which is a valid technique. However, there are several critical bugs in the implementation:

1. Collision handling is broken: The double hashing approach uses key % buckets for primary hash and key / bucketItems for secondary hash. This doesn't properly handle collisions. For example, key=1000 and key=2000 would both hash to primary index 0, but key=1000 has secondary index 1 and key=2000 has secondary index 2. So they wouldn't collide in the array, but key=1 and key=1001 would both have primary index 1 and secondary index 0, causing a collision that isn't handled.

2. The sentinel value approach is flawed: The student uses 0 as a sentinel for "not present" and -2 for "value is 0". But this fails when:

   • put(5, 0) is called, then get(5) should return 0
   • But the code stores -2 in the array, and get checks for -2 to return 0
   • However, if remove(5) is called, it sets the value to -1, but get doesn't check for -1, it only checks for 0
   • Wait, let me re-read... remove sets to -1, but get checks arr[firstIndex][secondIndex] == 0 to return -1. So if we remove a key, the value is -1, and get would return -1 (since -1 != 0), which is wrong - it should return -1 because the key was removed, but the code would return -1 anyway... actually this might work coincidentally.

Let me trace through more carefully:

• put(1, 1): firstIndex=1, secondIndex=0. arr[1] is null, so create arr[1] = new int[1000]. arr[1][0] = 1.
• get(1): firstIndex=1, secondIndex=0. arr[1] is not null, arr[1][0] = 1, not 0, not -2. Return 1. ✓
• put(2, 2): firstIndex=2, secondIndex=0. arr[2] is null, create arr[2]. arr[2][0] = 2.
• get(3): firstIndex=3, secondIndex=0. arr[3] is null. Return -1. ✓
• put(2, 1): firstIndex=2, secondIndex=0. arr[2] not null. arr[2][0] = 1. ✓
• get(2): firstIndex=2, secondIndex=0. arr[2][0] = 1. Return 1. ✓
• remove(2): firstIndex=2, secondIndex=0. arr[2][0] = -1.
• get(2): firstIndex=2, secondIndex=0. arr[2][0] = -1, not 0, not -2. Return -1. ✓

Now let's check the collision case:

• put(1001, 5): firstIndex = 1001 % 1000 = 1, secondIndex = 1001 / 1000 = 1. arr[1][1] = 5.
• put(1, 1): firstIndex = 1, secondIndex = 0. arr[1][0] = 1.
• get(1001): firstIndex = 1, secondIndex = 1. arr[1][1] = 5. Return 5. ✓
• get(1): firstIndex = 1, secondIndex = 0. arr[1][0] = 1. Return 1. ✓

Wait, but what about key=1001 and key=2001?

• put(1001, 5): firstIndex = 1, secondIndex = 1. arr[1][1] = 5.
• put(2001, 10): firstIndex = 2001 % 1000 = 1, secondIndex = 2001 / 1000 = 2. arr[1][2] = 10.
• get(1001): firstIndex = 1, secondIndex = 1. arr[1][1] = 5. Return 5. ✓
• get(2001): firstIndex = 1, secondIndex = 2. arr[1][2] = 10. Return 10. ✓

Hmm, actually the double hashing might work for distinct keys because the combination of primary and secondary hash is unique for each key... wait, no. Consider key=1 and key=1001:

• key=1: firstIndex=1, secondIndex=0
• key=1001: firstIndex=1, secondIndex=1

These are different positions, so no collision. But what about key=1000 and key=2000?

• key=1000: firstIndex=0, secondIndex=1
• key=2000: firstIndex=0, secondIndex=2

Different positions. What about key=500 and key=1500?

• key=500: firstIndex=500, secondIndex=0
• key=1500: firstIndex=500, secondIndex=1

Different positions.

Actually, I think the double hashing approach here might work because for any two distinct keys, the pair (primary hash, secondary hash) is unique. Let me verify:

• If key1 != key2, then either key1 % buckets != key2 % buckets, or key1 / bucketItems != key2 / bucketItems.
• Actually, this isn't necessarily true. Consider key1=1 and key2=1001: both have primary hash 1, but secondary hashes are 0 and 1 respectively. So they're at different positions.
• Consider key1=1001 and key2=2001: both have primary hash 1, but secondary hashes are 1 and 2 respectively. Different positions.

Actually, I think for any two distinct keys, the pair (key % buckets, key / bucketItems) is unique. This is because:

• If key1 % buckets == key2 % buckets and key1 / bucketItems == key2 / bucketItems, then key1 and key2 have the same remainder when divided by buckets and the same quotient when divided by bucketItems. This means key1 = key2 (since key = quotient * bucketItems + remainder, and bucketItems >= buckets).

Wait, bucketItems = 1000 and buckets = 1000. So key = (key / 1000) * 1000 + (key % 1000). If two keys have the same quotient and remainder, they must be equal. So yes, the double hashing approach is correct in terms of avoiding collisions!

But wait, there's still the issue with the sentinel values. Let me check:

• put(5, 0): firstIndex=5, secondIndex=0. arr[5][0] = -2.
• get(5): firstIndex=5, secondIndex=0. arr[5][0] = -2. Return 0. ✓
• remove(5): firstIndex=5, secondIndex=0. arr[5][0] = -1.
• get(5): firstIndex=5, secondIndex=0. arr[5][0] = -1, not 0, not -2. Return -1. ✓

But what about:

• put(5, 0): arr[5][0] = -2.
• put(5, 1): arr[5][0] = 1.
• get(5): arr[5][0] = 1. Return 1. ✓

What about:

• put(5, 0): arr[5][0] = -2.
• remove(5): arr[5][0] = -1.
• put(5, 0): arr[5][0] = -2.
• get(5): arr[5][0] = -2. Return 0. ✓

Hmm, but what about the case where value is -1 or -2?

• put(5, -1): value != 0, so arr[5][0] = -1.
• remove(5): arr[5][0] = -1 (already -1, no change).
• get(5): arr[5][0] = -1, not 0, not -2. Return -1. ✓ (because it was removed)

But what if we never remove and just put -1?

• put(5, -1): arr[5][0] = -1.
• get(5): arr[5][0] = -1, not 0, not -2. Return -1. ✗ (should return -1)

Wait, the constraints say 0 <= key, value <= 10^6, so value is always >= 0. So we don't need to worry about negative values being put.

But there's still a subtle issue: what if we put a value of 0, then remove it, then put a value of 0 again?

• `put(5,

VERDICT: NEEDS_IMPROVEMENT