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

Recall & Review

beginner

What is collision in a hash table?

Collision happens when two different keys hash to the same index in the hash table.

beginner

What does 'open addressing' mean in collision handling?

Open addressing means finding another empty spot in the hash table when a collision occurs, instead of using linked lists.

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 slot to place the new key.

intermediate

What problem can linear probing cause if many collisions happen?

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

beginner

In linear probing, how do you find a key during search?

Start at the hashed index and check slots one by one until you find the key or an empty slot (meaning key not present).

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

A Discard the new key

B Create a linked list at the collided slot

C Insert the key in the next available slot sequentially

D Resize the hash table immediately

Linear probing inserts the key in the next available slot by checking slots one by one.

Which of these is a downside of linear probing?

A It can cause clustering of keys

B It never handles collisions

C It uses extra memory for linked lists

D It requires complex data structures

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

How do you know a key is not in the hash table when searching with linear probing?

A When the table is full

B When you reach the end of the table

C When you find a different key

D When you find an empty slot during probing

Finding an empty slot during probing means the key was never inserted.

What is the first step in inserting a key using linear probing?

A Search the entire table

B Hash the key to get an index

C Resize the table

D Delete an existing key

You first hash the key to find the initial index to try inserting.

If the hashed index is occupied, what does linear probing do next?

A Check the next index (index + 1)

B Stop and fail insertion

C Jump to a random index

D Use a linked list at that index

Linear probing checks the next index sequentially until it finds an empty slot.

Explain how collision handling works using open addressing with linear probing.

Think about what happens when two keys want the same spot and how linear probing finds a new spot.

Describe the main advantage and disadvantage of using linear probing for collision handling.

Consider memory use and performance impact.

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 to store multiple elements at the same index.

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

D. Rehash all elements using a new hash function when a collision occurs.

Solution

1. Step 1: Understand linear probing concept

   Linear probing handles collisions by checking the next slot in the array sequentially when the desired slot is occupied.

2. Step 2: Compare options with definition

   If a collision occurs, check the next slot sequentially until an empty slot is found. correctly describes this sequential checking. Other options describe different collision handling methods.

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 rehashing happens on every collision
• Assuming table size doubles automatically

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. next_index = (i + 3) % m

B. next_index = (i * 2) % m

C. next_index = (i - 1) % m

D. next_index = (i + 1) % m

Solution

1. Step 1: Recall linear probing formula

   Linear probing moves to the next slot by adding 1 to the current index and taking modulo with table size.

2. Step 2: Match formula with options

   next_index = (i + 1) % m matches this formula: (i + 1) % m. Other options use incorrect increments.

3. Final Answer:

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

4. Quick Check:

   Linear probing next index = (i + 1) mod m [OK]

Hint: Linear probing adds 1 to index modulo table size [OK]

Common Mistakes:

• Using multiplication instead of addition
• Subtracting instead of adding
• Adding a number other than 1

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

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

Solution

1. Step 1: Calculate initial positions using hash function

   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, indices 2 and 3 are occupied, so insert at index 4.

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Placing all keys at same index ignoring collisions
• Not wrapping around table size
• Misplacing keys at wrong indices

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

int hash(int key, int size) { return key % size; }
void insert(int table[], int size, int key) {
  int index = hash(key, size);
  while (table[index] != -1) {
    index = (index + 1) % size;
  }
  table[index] = key;
}

medium

A. The index update should subtract 1 instead of adding 1.

B. The table is not initialized with -1 for empty slots.

C. The while loop condition should check for table[index] == -1.

D. The hash function should use multiplication instead of modulo.

Solution

1. Step 1: Analyze the insert function logic

   The function checks for empty slots by comparing with -1, so the table must be initialized with -1 to mark empty slots.

2. Step 2: Identify missing initialization

   If the table is not initialized with -1, the condition table[index] != -1 may fail incorrectly, causing wrong insertions or infinite loops.

3. Final Answer:

   The table is not initialized with -1 for empty slots. -> Option B

4. Quick Check:

   Empty slots must be marked -1 [OK]

Hint: Initialize table slots with -1 before insertion [OK]

Common Mistakes:

• Ignoring table initialization
• Wrong loop condition in while
• Incorrect hash function usage

5. You have a hash table of size 7 using linear probing. Insert keys 10, 20, 15, 7, 32 in order using hash function h(k) = k % 7. After all insertions, what is the content of the table?

hard

A. [7, 15, null, 10, 32, null, 20]

B. [null, 15, 32, 10, null, 20, 7]

C. [null, 15, 32, 10, 20, 7, null]

D. [null, 15, 32, 10, 20, null, 7]

Solution

1. Step 1: Calculate initial indices for each key

   h(10)=3, h(20)=6, h(15)=1, h(7)=0, h(32)=4.

2. Step 2: Insert keys using linear probing

   Insert 10 at index 3. Insert 20 at index 6. Insert 15 at index 1. Insert 7 at index 0. Insert 32 at index 4 (empty). Final table: [7, 15, null, 10, 32, null, 20].

3. Step 3: Match with options

   [7, 15, null, 10, 32, null, 20] matches the correct order. Nulls at indices 2 and 5.

4. Final Answer:

   [7, 15, null, 10, 32, null, 20] -> Option A

5. Quick Check:

   Keys placed at their hash or next free slot [OK]

Hint: Calculate hash, insert or probe next free slot [OK]

Common Mistakes:

• Not probing when slot is occupied
• Misplacing keys at wrong indices
• Ignoring table size wrap-around