Why Collision Handling Using Open Addressing Linear Probing in DSA Typescript?

The Big Idea

What if your data suddenly disappears because two items wanted the same spot?

The Scenario

Imagine you have a small mailbox with numbered slots for letters. When two letters arrive for the same slot number, you have to decide where to put the second letter. Without a plan, you might just randomly place it somewhere else, making it hard to find later.

The Problem

Manually searching for a free slot when a collision happens is slow and confusing. You might overwrite letters or lose track of where you put them. This causes mistakes and wastes time when you want to find a letter quickly.

The Solution

Open Addressing with Linear Probing solves this by checking the next slot in order until it finds an empty one. This way, all letters are stored in a predictable sequence, making it easy to find them later without extra storage.

Before vs After

✗ Before

function insert(key) {
  let index = hash(key);
  if (table[index] !== null) {
    // no plan, just put somewhere random
    index = findRandomEmptySlot();
  }
  table[index] = key;
}

✓ After

function insert(key) {
  let index = hash(key);
  while (table[index] !== null) {
    index = (index + 1) % table.length;
  }
  table[index] = key;
}

What It Enables

This method lets us store and find data quickly and reliably, even when many items want the same spot.

Real Life Example

Think of a parking lot with numbered spaces. If your spot is taken, you look for the next free space in order. This keeps parking organized and easy to manage.

Key Takeaways

Manual collision handling is slow and error-prone.

Linear probing checks next slots in order to find free space.

This keeps data organized and easy to find.

Practice

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

easy

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

B. Use a linked list to store multiple values at the same index.

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

D. Replace the existing value at the collided slot with the new value.

Solution

1. Step 1: Understand collision handling

   When two keys hash to the same index, a collision occurs.

2. Step 2: Explain linear probing method

   Linear probing resolves collisions by checking the next slot in order until an empty slot is found.

3. Final Answer:

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

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 on collision

2. Which of the following TypeScript code snippets correctly implements the linear probing step to find the next index after a collision in a hash table of size size?

easy

A. let nextIndex = currentIndex % size;

B. let nextIndex = (currentIndex * 2) % size;

C. let nextIndex = (currentIndex - 1) % size;

D. let nextIndex = (currentIndex + 1) % size;

Solution

1. Step 1: Understand linear probing index calculation

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

2. Step 2: Check code correctness

   let nextIndex = (currentIndex + 1) % size; correctly calculates nextIndex as (currentIndex + 1) % size, ensuring wrap-around.

3. Final Answer:

   let nextIndex = (currentIndex + 1) % size; -> Option D

4. Quick Check:

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

Hint: Add 1 and mod by size for next slot [OK]

Common Mistakes:

• Multiplying index instead of adding
• Subtracting index causing negative values
• Not using modulo for wrap-around

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

medium

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

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

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

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

Solution

1. Step 1: Calculate initial hash indices

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

2. Step 2: Insert keys using linear probing

   Insert 7 at index 2. For 12, index 2 is occupied, try 3 (empty), insert 12. For 17, indices 2 and 3 occupied, try 4 (empty), insert 17.

3. Final Answer:

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

4. Quick Check:

   All keys start at 2, linear probe fills next slots [OK]

Hint: All keys hash same, fill next slots sequentially [OK]

Common Mistakes:

• Placing all keys at index 2 ignoring collisions
• Not wrapping around the table
• Misplacing keys at wrong indices

4. The following TypeScript code tries to insert keys into a hash table using linear probing. What is the bug that causes an infinite loop?

const size = 5;
const table = new Array(size).fill(null);
function hash(key: number) { return key % size; }
function insert(key: number) {
  let index = hash(key);
  while (table[index] !== null) {
    index = (index + 1) % size;
  }
  table[index] = key;
}
insert(5);
insert(10);
insert(15);
insert(20);
insert(25);
insert(30);

medium

A. No check for full table causing infinite loop when table is full.

B. Hash function is incorrect and returns negative indices.

C. Array initialization with null is wrong; should use undefined.

D. Insert function does not update index inside the loop.

Solution

1. Step 1: Analyze insert loop behavior

   The while loop keeps moving index until an empty slot is found.

2. Step 2: Identify missing full table check

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

3. Final Answer:

   No check for full table causing infinite loop when table is full. -> Option A

4. Quick Check:

   Full table without exit causes infinite loop [OK]

Hint: Add full table check to avoid infinite loop [OK]

Common Mistakes:

• Assuming hash function returns negative
• Thinking null vs undefined causes loop
• Forgetting to update index inside loop

5. You have a hash table of size 7 using linear probing. The hash function is h(key) = key % 7. You insert keys: 14, 21, 28, 35, 42. Then you delete key 21. What is the correct way to handle deletion to avoid search errors in linear probing?

hard

A. Set the slot to null immediately to free it for new insertions.

B. Mark the slot as a special 'deleted' marker instead of null to continue probing.

C. Rehash the entire table after deletion to fix collisions.

D. Do nothing; deletion is not needed in linear probing.

Solution

1. Step 1: Understand deletion problem in linear probing

   Setting slot to null breaks the probe chain, causing search failures for keys after deleted slot.

2. Step 2: Use special 'deleted' marker

   Marking slot as 'deleted' allows probing to continue while allowing new insertions.

3. Final Answer:

   Mark the slot as a special 'deleted' marker instead of null to continue probing. -> Option B

4. Quick Check:

   Deleted marker keeps probe chain intact [OK]

Hint: Use 'deleted' marker slot, not null, after deletion [OK]

Common Mistakes:

• Setting slot to null breaks search chain
• Rehashing entire table unnecessarily
• Ignoring deletion impact on probing