Collision Handling Using Open Addressing Linear Probing in DSA Typescript - Time & Space Complexity

Time Complexity: Collision Handling Using Open Addressing Linear Probing

O(n)

Understanding Time Complexity

When we use open addressing with linear probing to handle collisions in a hash table, we want to know how the time to insert or find an item changes as the table fills up.

We ask: How does the number of steps grow when the table gets more crowded?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function linearProbingInsert(table: (number | null)[], key: number): void {
  const size = table.length;
  let index = key % size;
  while (table[index] !== null) {
    index = (index + 1) % size;
  }
  table[index] = key;
}
    

This code inserts a key into a hash table using linear probing to find the next empty slot if a collision occurs.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that checks each slot until an empty one is found.
• How many times: In the worst case, it can check every slot once if the table is almost full.

How Execution Grows With Input

As the table fills, the number of checks grows because more slots are occupied, so the loop runs longer.

Input Size (n)	Approx. Operations
10 (mostly empty)	1-2 checks
100 (half full)	Up to 50 checks
1000 (almost full)	Up to 1000 checks

Pattern observation: The number of checks grows roughly proportional to how full the table is, so more items mean more steps.

Final Time Complexity

Time Complexity: O(n)

This means in the worst case, the time to insert or find an item grows linearly with the number of items in the table.

Common Mistake

[X] Wrong: "Insertion always takes constant time because hashing is fast."

[OK] Correct: When many collisions happen, linear probing may check many slots, making insertion slower as the table fills.

Interview Connect

Understanding how collision handling affects time helps you explain trade-offs in hash tables clearly, a skill valued in real-world coding and interviews.

Self-Check

"What if we changed linear probing to quadratic probing? How would the time complexity change?"

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