Collision Handling Using Open Addressing Linear Probing in DSA Java - 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. This means both want to be stored in the same spot.

beginner

What is open addressing in hash tables?

Open addressing is a way to handle collisions by finding another empty spot in the table instead of using linked lists or extra structures.

beginner

How does linear probing work in open addressing?

Linear probing checks the next slot in the table one by one until it finds an empty spot to store the new key.

intermediate

What problem can occur if the hash table is almost full when using linear probing?

Clustering happens, where many keys group together, causing longer search times because many slots must be checked.

intermediate

Why is it important to use a good hash function with linear probing?

A good hash function spreads keys evenly to reduce collisions and clustering, making linear probing efficient.

What does linear probing do when a collision occurs?

A Checks the next slot sequentially until an empty one is found

B Creates a linked list at the collided slot

C Rehashes the entire table

D Deletes the existing key

Linear probing searches the next slots one by one until it finds an empty slot.

In open addressing, what happens if the table is full and no empty slot is found during insertion?

A The table automatically resizes

B Insertion fails because no space is available

C The key replaces an existing key

D The key is stored outside the table

If no empty slot is found, insertion fails because the table is full.

Which of these is a disadvantage of linear probing?

A Uses extra memory for linked lists

B Requires complex data structures

C Clustering of keys causing slower searches

D Cannot handle collisions

Linear probing can cause clustering, where many keys group together, slowing down operations.

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

A To store all keys without losing any data

B To delete duplicate keys

C To sort keys in order

D To compress the data

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

Which method is NOT a form of open addressing?

A Linear probing

B Quadratic probing

C Double hashing

D Separate chaining

Separate chaining uses linked lists, not open addressing.

Explain how linear probing handles collisions in a hash table.

Think about what happens when two keys want the same spot.

Describe the advantages and disadvantages of using linear probing for collision handling.

Consider both ease and performance.

Practice

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

easy

A. If a slot is occupied, check the next slot sequentially until an empty one is found.

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

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

D. Rehash all elements using a new hash function.

Solution

1. Step 1: Understand linear probing concept

   Linear probing means when a collision happens, we look at the next slot in order, moving forward one by one.

2. Step 2: Compare options

   If a slot is occupied, check the next slot sequentially until an empty one is found. describes this exact process. Options B, C, and D describe other collision methods or resizing, not linear probing.

3. Final Answer:

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

4. Quick Check:

   Linear probing = sequential slot check [OK]

Hint: Linear probing checks next slots one by one [OK]

Common Mistakes:

• Confusing linear probing with chaining
• Thinking linear probing resizes the table
• Assuming linear probing uses a new hash function

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 * 2) % m

D. (i - 1) % m

Solution

1. Step 1: Recall linear probing index calculation

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

2. Step 2: Check options

   Only (i + 1) % m uses (i + 1) % m, which correctly moves to the next slot circularly.

3. Final Answer:

   (i + 1) % m -> Option B

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:

• Adding 2 instead of 1
• Multiplying index instead of adding
• Subtracting index causing negative values

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?

medium

A. [7, 12, 17, null, null]

B. [7, null, 12, 17, null]

C. [7, 12, null, 17, null]

D. [null, null, 7, 12, 17]

Solution

1. Step 1: Calculate initial positions

   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 occupied, so try index 3. Insert 12 at index 3. For 17, indexes 2 and 3 are occupied, so insert at index 4.

3. Final Answer:

   [null, null, 7, 12, 17] -> Option D

4. Quick Check:

   Keys placed at 2, 3, 4 sequentially [OK]

Hint: Check hash, then next slots if occupied [OK]

Common Mistakes:

• Placing all keys at initial hash index
• Not wrapping around the table
• Confusing index positions

4. Given the following code snippet for linear probing insertion, what is the bug?

int insert(int[] table, int key) {
  int m = table.length;
  int i = key % m;
  while (table[i] != 0) {
    i = (i + 1) % m;
  }
  table[i] = key;
  return i;
}

medium

A. It does not handle the case when the table is full, causing an infinite loop.

B. The hash function should be key / m instead of key % m.

C. It inserts the key at the wrong index without probing.

D. It uses 0 to check empty slots, which is invalid for all keys.

Solution

1. Step 1: Analyze the while loop condition

   The loop continues while the slot is occupied (not zero). If the table is full, it loops forever because no empty slot is found.

2. Step 2: Check other options

   The hash function should be key / m instead of key % m. is wrong because modulo is correct for hashing. It inserts the key at the wrong index without probing. is wrong because probing is done. It uses 0 to check empty slots, which is invalid for all keys. assumes 0 is invalid, but 0 can be used as empty marker if keys are positive.

3. Final Answer:

   It does not handle the case when the table is full, causing an infinite loop. -> Option A

4. Quick Check:

   Full table causes infinite loop [OK]

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

Common Mistakes:

• Ignoring full table condition
• Misunderstanding hash function usage
• Assuming 0 cannot be empty slot marker

5. You have a hash table of size 7 using linear probing. The hash function is h(k) = k % 7. Insert keys 10, 24, 31, 17, 3, 7 in order. After all insertions, which index will key 7 be stored at?

hard

A. 3

B. 4

C. 1

D. 6

Solution

1. Step 1: Calculate initial hashes and insert keys

   h(10)=3, insert at 3.
   h(24)=3, slot 3 occupied, try 4.
   h(31)=3, slots 3 and 4 occupied, try 5.
   h(17)=3, slots 3,4,5 occupied, try 6.
   h(3)=3, slots 3,4,5,6 occupied, try 0.
   h(7)=0, slot 0 occupied, try 1.

2. Step 2: Check slot 1 for key 7

   Slot 1 is empty, so key 7 is stored at index 1.

3. Final Answer:

   1 -> Option C

4. Quick Check:

   Linear probing finds next free slot [OK]

Hint: Probe next slots until empty to find key position [OK]

Common Mistakes:

• Stopping at initial hash index
• Not checking all occupied slots
• Miscounting index after wrap-around