Collision handling (open addressing) in Data Structures Theory - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to identify the method used in open addressing for collision handling.

Data Structures Theory

In open addressing, when a collision occurs, the algorithm tries to find the next [1] slot in the hash table.

Drag options to blanks, or click blank then click option'

Aempty

Boccupied

Cfull

Dlocked

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing 'occupied' or 'full' because they sound similar to 'empty'.

Open addressing searches for the next empty slot to resolve collisions.

2fill in blank

medium

Complete the code to describe the probing sequence in linear probing.

Data Structures Theory

In linear probing, the next slot is found by moving [1] step(s) forward from the current slot.

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Atwo

Bone

Cthree

Dzero

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing 'two' or 'three' thinking it jumps multiple slots.

Linear probing moves one step forward to find the next slot.

3fill in blank

hard

Fix the error in the description of quadratic probing.

Data Structures Theory

Quadratic probing finds the next slot by adding [1] to the original hash index, where i is the probe number.

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Ai + 1

Bi - 1

C2 * i

Di * i

Attempts:

3 left

💡 Hint

Common Mistakes

Using linear increments like 'i + 1' instead of quadratic increments.

Quadratic probing adds i squared (i * i) to the original hash index to find the next slot.

4fill in blank

hard

Fill both blanks to complete the formula for double hashing.

Data Structures Theory

The double hashing formula is: (hash1(key) + [1] * [2]) mod table_size.

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Ai

Bhash2(key)

Ckey

Dtable_size

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'key' or 'table_size' instead of the second hash function.

Double hashing uses the formula: (hash1(key) + i * hash2(key)) mod table_size to find the next slot.

5fill in blank

hard

Fill all three blanks to complete the explanation of open addressing.

Data Structures Theory

In open addressing, all elements are stored in the [1] itself. When a collision occurs, the algorithm probes the table using a sequence defined by [2] until an [3] slot is found.

Drag options to blanks, or click blank then click option'

Ahash table

Bprobing function

Cempty

Dlinked list

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing 'linked list' for storage instead of 'hash table'.

Confusing 'probing function' with 'empty' slot.

Open addressing stores all elements in the hash table itself. It uses a probing function to find the next slot until an empty slot is found.

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