Collision Handling Using Open Addressing Linear Probing in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is collision in a hash table?

Collision happens when two different keys get the same index in the hash table. It means they want to sit in the same chair at the same time.

beginner

What does open addressing mean in collision handling?

Open addressing means when a collision happens, we look for another empty spot in the table to put the new key instead of using extra space like linked lists.

beginner

How does linear probing find the next spot after a collision?

Linear probing checks the next spots one by one in order (like moving to the next chair) until it finds an empty spot to place the key.

intermediate

What is a drawback of linear probing?

Linear probing can cause clustering, where many keys group together, making it slower to find empty spots or keys because you have to check many spots in a row.

intermediate

In linear probing, how do you calculate the next index to check?

You add 1 to the current index and take the remainder when divided by the table size (nextIndex = (currentIndex + 1) % tableSize). This wraps around to the start if needed.

What happens when a collision occurs in open addressing with linear probing?

A Look for the next empty spot by checking indexes one by one

B Create a linked list at the collided index

C Ignore the new key

D Resize the hash table immediately

Linear probing searches the next spots one by one to find an empty place.

Which of these is a problem caused by linear probing?

A Keys are stored in random order

B Clustering of keys in the table

C Keys never collide

D Using extra memory for linked lists

Linear probing can cause clustering, where keys group together making search slower.

How do you calculate the next index in linear probing?

A currentIndex - 1

B currentIndex * 2

C hash(key) + 5

D (currentIndex + 1) % tableSize

The next index is the current index plus one, wrapped around by the table size.

What is the main goal of collision handling in hash tables?

A To store keys without losing any data

B To delete keys quickly

C To sort keys alphabetically

D To reduce the size of the table

Collision handling ensures all keys can be stored even if they map to the same index.

In open addressing, what do you do if the table is full and you cannot find an empty spot?

A Overwrite existing keys

B Stop inserting new keys

C Resize the table and rehash all keys

D Ignore the new key

When the table is full, resizing and rehashing is needed to insert new keys.

Explain how linear probing handles collisions in a hash table.

Think about moving step-by-step to find a free chair.

What are the advantages and disadvantages of using linear probing for collision handling?

Consider both the easy parts and the problems it can cause.

Practice

1. What is the main idea behind linear probing in collision handling for hash tables?

easy

A. If a collision occurs, check the next slot sequentially until an empty slot is found.

B. Use a linked list at each slot to store multiple values.

C. Double the size of the hash table when a collision happens.

D. Replace the existing value in the slot with the new value.

Solution

1. Step 1: Understand collision handling methods

   Linear probing handles collisions by searching the next available slot in a sequence.

2. Step 2: Identify linear probing behavior

   When a collision happens, linear probing checks the next slot one by one until it finds an empty slot.

3. Final Answer:

   If a collision occurs, check the next slot sequentially until an empty slot is found. -> Option A

4. Quick Check:

   Linear probing = sequential slot check [OK]

Hint: Linear probing means try next slot one by one [OK]

Common Mistakes:

• Confusing linear probing with chaining
• Thinking collision means resizing immediately
• Replacing existing values instead of probing

2. Which of the following is the correct way to calculate the next index in linear probing for a hash table of size m when current index is i?

easy

A. (i * 2) % m

B. (i - 1) % m

C. (i + 3) % m

D. (i + 1) % m

Solution

1. Step 1: Recall linear probing index update

   Linear probing moves to the next slot by adding 1 to the current index.

2. Step 2: Apply modulo to wrap around

   To stay within table size, use modulo operation with table size m.

3. Final Answer:

   (i + 1) % m -> Option D

4. Quick Check:

   Next index = (current + 1) mod size [OK]

Hint: Next index in linear probing is (i + 1) mod table size [OK]

Common Mistakes:

• Using multiplication instead of addition
• Not using modulo to wrap index
• Subtracting instead of adding

3. Consider a hash table of size 5 using linear probing. The hash function is h(k) = k % 5. Insert keys 7, 12, 17 in order. What is the final state of the table (index: value)?

medium

A. 0: null, 1: null, 2: 7, 3: 12, 4: 17

B. 0: null, 1: 7, 2: null, 3: 12, 4: 17

C. 0: 7, 1: 12, 2: 17, 3: null, 4: null

D. 0: null, 1: 7, 2: 12, 3: 17, 4: null

Solution

1. Step 1: Calculate initial positions using hash function

   7 % 5 = 2, 12 % 5 = 2, 17 % 5 = 2. All keys initially map to index 2.

2. Step 2: Insert keys using linear probing

   Insert 7 at index 2 (empty). Insert 12: index 2 occupied, try 3 (empty), insert at 3. Insert 17: index 2 occupied, 3 occupied, try 4 (empty), insert at 4.

3. Final Answer:

   0: null, 1: null, 2: 7, 3: 12, 4: 17 -> Option A

4. Quick Check:

   Keys placed at 2, 3, 4 with probing [OK]

Hint: All keys hash to 2, probe next slots for empty [OK]

Common Mistakes:

• Placing all keys at index 2 ignoring collisions
• Skipping slots incorrectly during probing
• Miscomputing hash values

4. The following code snippet tries to insert keys into a hash table using linear probing. What is the bug?

const size = 7;
const table = new Array(size).fill(null);
function hash(key) {
  return key % size;
}
function insert(key) {
  let index = hash(key);
  while (table[index] !== null) {
    index = (index + 1) % size;
  }
  table[index] = key;
}
insert(10);
insert(17);
insert(24);
console.log(table);

medium

A. The hash function should use key % (size - 1) instead of key % size.

B. The table array should be initialized with zeros instead of null.

C. The insert function does not handle the case when the table is full, causing an infinite loop.

D. The index update in the while loop should subtract 1 instead of adding 1.

Solution

1. Step 1: Analyze insert function loop

   The while loop probes next slots until it finds an empty slot (null).

2. Step 2: Identify missing full table check

   If the table is full, the loop cycles endlessly without exit condition, causing infinite loop.

3. Final Answer:

   The insert function does not handle the case when the table is full, causing an infinite loop. -> Option C

4. Quick Check:

   Missing full table check causes infinite loop [OK]

Hint: Always check for full table to avoid infinite loops [OK]

Common Mistakes:

• Thinking modulo should be size-1
• Ignoring infinite loop possibility
• Confusing null with zero initialization

5. You have a hash table of size 10 using linear probing. The table currently stores keys at indices: 2, 3, 4, 5. You want to insert key 22. The hash function is h(k) = k % 10. Which index will key 22 be inserted into?

hard

A. 2

B. 6

C. 7

D. 5

Solution

1. Step 1: Calculate initial index for key 22

   22 % 10 = 2, so initial index is 2.

2. Step 2: Apply linear probing to find empty slot

   Indices 2, 3, 4, 5 are occupied. Next index is 6, which is free.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Start at 2, probe until free slot at 6 [OK]

Hint: Probe forward from hash index until empty slot found [OK]

Common Mistakes:

• Inserting at first hashed index ignoring occupancy
• Stopping probing too early
• Choosing wrong next index