Collision handling (open addressing) in Data Structures Theory - Time & Space Complexity

Time Complexity: Collision handling (open addressing)

O(n)

Understanding Time Complexity

When storing data in a hash table, sometimes two items want the same spot. Open addressing is one way to handle this by finding another spot.

We want to know how the time to find or insert items grows as the table fills up.

Scenario Under Consideration

Analyze the time complexity of inserting an item using linear probing in open addressing.

function insert(key, table) {
  let index = hash(key) % table.size;
  while (table[index] !== undefined) {
    index = (index + 1) % table.size;  // move to next slot
  }
  table[index] = key;
}
    

This code tries to place a key in the hash table. If the spot is taken, it checks the next spot until it finds an empty one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Checking table slots one by one (linear probing).
• How many times: Up to the number of slots checked until an empty one is found, which depends on how full the table is.

How Execution Grows With Input

As the table fills, more slots may be checked before finding an empty one.

Input Size (n)	Approx. Operations
10 (mostly empty)	About 1 to 2 checks
100 (half full)	Several checks, maybe 5 to 10
1000 (almost full)	Many checks, possibly hundreds

Pattern observation: The more full the table, the longer it takes to find an empty slot, so time grows as the load increases.

Final Time Complexity

Time Complexity: O(n)

This means in the worst case, you might check many slots one by one, so time grows linearly with the number of items.

Common Mistake

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

[OK] Correct: When many slots are filled, you may need to check many places before finding an empty one, making insertion slower.

Interview Connect

Understanding how open addressing affects time helps you explain trade-offs in hash tables clearly and shows you grasp real-world data structure behavior.

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 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