Collision handling (open addressing) in Data Structures Theory - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is collision handling in open addressing?

Collision handling in open addressing is a method to resolve conflicts when two keys hash to the same index by finding another open slot within the hash table itself.

beginner

Name three common probing techniques used in open addressing.

The three common probing techniques are linear probing, quadratic probing, and double hashing.

beginner

How does linear probing work in open addressing?

Linear probing checks the next slot sequentially (index + 1, index + 2, etc.) until it finds an empty slot to place the new key.

intermediate

What problem can occur with linear probing and how is it called?

Linear probing can cause clustering, where groups of filled slots form, increasing search time. This is called primary clustering.

intermediate

Explain double hashing in open addressing.

Double hashing uses a second hash function to calculate the step size for probing, reducing clustering by spreading out keys more evenly.

What does open addressing do when a collision occurs?

A Finds another empty slot in the hash table

B Stores the key in a linked list

C Deletes the existing key

D Ignores the new key

Open addressing resolves collisions by searching for another empty slot within the hash table.

Which probing method checks slots sequentially after a collision?

A Linear probing

B Double hashing

C Quadratic probing

D Separate chaining

Linear probing checks the next slots one by one in order.

What is the main disadvantage of linear probing?

A Slow hash function

B Secondary clustering

C High memory usage

D Primary clustering

Linear probing can cause primary clustering, where filled slots group together and slow down operations.

Which technique uses a second hash function to find the next slot?

A Quadratic probing

B Linear probing

C Double hashing

D Separate chaining

Double hashing uses a second hash function to calculate the step size for probing.

In open addressing, what happens if the table is full?

A Keys are stored outside the table

B Insertion fails because no empty slot is available

C The table automatically expands

D Old keys are overwritten

If the hash table is full, open addressing cannot insert new keys because no empty slots remain.

Describe how open addressing handles collisions in a hash table.

Think about how the table itself is used to resolve conflicts.

Compare linear probing and double hashing as collision handling methods.

Focus on how they find the next slot and their impact on clustering.

Practice

1. What is the main idea behind open addressing in collision handling?

easy

A. Find another empty spot in the hash table when a collision occurs

B. Store collided elements in a separate linked list

C. Use multiple hash functions to avoid collisions

D. Resize the hash table immediately after a collision

Solution

1. Step 1: Understand collision handling methods

   Open addressing handles collisions by searching for another empty slot inside the hash table itself.

2. Step 2: Compare with other methods

   Unlike chaining (which uses linked lists), open addressing keeps all data inside the array by probing for free spots.

3. Final Answer:

   Find another empty spot in the hash table when a collision occurs -> Option A

4. Quick Check:

   Open addressing = find empty spot inside table [OK]

Hint: Open addressing means searching inside the table for free spots [OK]

Common Mistakes:

• Confusing open addressing with chaining
• Thinking collisions are ignored
• Assuming resizing happens immediately on collision

2. Which of the following is the correct syntax to start linear probing from index i in a hash table of size m?

easy

A. next_index = (i - 1) % m

B. next_index = (i * 2) % m

C. next_index = (i + 1) % m

D. next_index = (i + m) % 2

Solution

1. Step 1: Recall linear probing formula

   Linear probing moves to the next slot by adding 1 to the current index and wrapping around using modulo.

2. Step 2: Apply modulo with table size

   To stay within the table size, use (i + 1) % m, where m is the table size.

3. Final Answer:

   next_index = (i + 1) % m -> Option C

4. Quick Check:

   Linear probing = increment index by 1 modulo table size [OK]

Hint: Linear probing adds 1 and wraps with modulo table size [OK]

Common Mistakes:

• Using multiplication instead of addition
• Incorrect modulo divisor
• Subtracting instead of adding

3. Given a hash table of size 5 and a hash function h(k) = k % 5, if keys 7, 12, and 17 are inserted using linear probing, what will be the final index of key 17?

medium

A. 3

B. 2

C. 0

D. 4

Solution

1. Step 1: Calculate initial hash indices

   h(7) = 7 % 5 = 2, h(12) = 12 % 5 = 2, h(17) = 17 % 5 = 2.

2. Step 2: Insert keys using linear probing

   Insert 7 at index 2. For 12, index 2 is taken, so try 3 (free). Insert 12 at 3. For 17, indices 2 and 3 are taken, try 4 (free). Insert 17 at 4.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Linear probing moves to next free slot: 17 at index 4 [OK]

Hint: Check initial hash, then probe next slots until free [OK]

Common Mistakes:

• Assuming all keys go to same index
• Not probing after collision
• Miscomputing modulo

4. What is the main problem with this linear probing insertion code snippet?

table = [None]*5
key = 12
index = key % 5
while table[index] is not None:
    index = (index + 1) % 5
# Missing insertion step here

medium

A. It never inserts the key into the table

B. It causes an infinite loop even if the table has empty slots

C. It uses wrong modulo operation

D. It overwrites existing keys without checking

Solution

1. Step 1: Analyze the loop behavior

   The loop finds the next free slot by incrementing index until it finds None.

2. Step 2: Check for insertion after loop

   After the loop, the code does not insert the key into the found free slot, so the key is never stored.

3. Final Answer:

   It never inserts the key into the table -> Option A

4. Quick Check:

   Missing insertion step means key not stored [OK]

Hint: Always insert key after finding free slot [OK]

Common Mistakes:

• Forgetting to insert after probing
• Ignoring full table infinite loop
• Wrong modulo usage

5. In a hash table using open addressing with linear probing, what happens if the table becomes completely full and you try to insert a new key?

hard

A. The new key overwrites the oldest key in the table

B. The insertion fails because no empty slot is available

C. The table automatically resizes and rehashes all keys

D. The insertion loops infinitely without stopping

Solution

1. Step 1: Understand open addressing limits

   Open addressing requires an empty slot to insert a new key by probing.

2. Step 2: Consider full table scenario

   If the table is full, no empty slot exists, so insertion cannot proceed and must fail or trigger resizing externally.

3. Final Answer:

   The insertion fails because no empty slot is available -> Option B

4. Quick Check:

   Full table means no free slot, insertion fails [OK]

Hint: Full table means no free slot, insertion fails [OK]

Common Mistakes:

• Assuming automatic resizing always happens
• Thinking keys get overwritten silently
• Ignoring infinite loop possibility