Collision Handling Using Open Addressing Linear Probing in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the hash index for a given key.

DSA Python

def hash_function(key, size):
    return key [1] size

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

A+

B-

C*

D%

Attempts:

3 left

💡 Hint

Common Mistakes

Using + or - instead of modulo causes index out of range errors.

Using * can produce very large numbers not suitable for indexing.

The modulo operator (%) is used to ensure the hash index stays within the table size.

2fill in blank

medium

Complete the code to find the next index in linear probing.

DSA Python

def linear_probe(index, size):
    return (index [1] 1) % size

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

A+

B-

C*

D//

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction (-) moves backward, which is incorrect for linear probing.

Using multiplication (*) or floor division (//) does not move to the next slot.

In linear probing, we move to the next slot by adding 1 to the current index.

3fill in blank

hard

Fix the error in the insertion loop to handle collisions correctly.

DSA Python

def insert(hash_table, key, size):
    index = hash_function(key, size)
    start_index = index
    while hash_table[index] is not None:
        index = (index [1] 1) % size
        if index == start_index:
            return False  # Table full
    hash_table[index] = key
    return True

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

A*

B-

C+

D//

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction (-) can cause infinite loops or wrong slot checks.

Using multiplication or floor division is incorrect for index increment.

We add 1 to the index to move to the next slot in linear probing.

4fill in blank

hard

Fill both blanks to complete the search function using linear probing.

DSA Python

def search(hash_table, key, size):
    index = hash_function(key, size)
    start_index = index
    while hash_table[index] is not None and hash_table[index] != [1]:
        index = (index [2] 1) % size
        if index == start_index:
            return -1  # Not found
    if hash_table[index] == key:
        return index
    return -1

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

Akey

Bsize

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with size instead of key causes wrong search results.

Using subtraction (-) instead of addition (+) moves backward incorrectly.

We compare the slot with the key and move forward by adding 1 to index.

5fill in blank

hard

Fill all three blanks to complete the delete function using linear probing.

DSA Python

def delete(hash_table, key, size):
    index = hash_function(key, size)
    start_index = index
    while hash_table[index] is not None:
        if hash_table[index] == [1]:
            hash_table[index] = [2]  # Mark as deleted
            return True
        index = (index [3] 1) % size
        if index == start_index:
            return False  # Not found
    return False

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

Akey

BNone

C+

D-

E0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting the slot to None prevents searches from probing past the deleted slot, causing items to become unreachable.

Using subtraction (-) instead of addition (+) moves backward incorrectly.

We check if the slot has the key, mark it as 0 (deleted marker) to delete, and move forward by adding 1.

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 occurs.

B. If a slot is occupied, check the next slot sequentially until an empty one is found.

C. Use a linked list at each slot to store multiple elements.

D. Rehash all elements using a new hash function.

Solution

1. Step 1: Understand linear probing concept

   Linear probing handles collisions by checking the next slot in order if the original slot is taken.

2. Step 2: Compare with other methods

   Other methods like chaining use linked lists, resizing, or rehashing, which are different from linear probing.

3. Final Answer:

   If a slot is occupied, check the next slot sequentially until an empty one is found. -> Option B

4. Quick Check:

   Linear probing = sequential slot check [OK]

Hint: Linear probing means try next slot until free [OK]

Common Mistakes:

• Confusing linear probing with chaining
• Thinking linear probing resizes the table
• Assuming rehashing is part of linear probing

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 a collision occurs at index i?

easy

A. (i + 2) % m

B. (i * 2) % m

C. (i + 1) % m

D. (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: Apply modulo for wrap-around

   Using modulo ensures the index wraps around to the start if it reaches the end of the table.

3. Final Answer:

   (i + 1) % m -> Option C

4. Quick Check:

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

Hint: Next slot = (current index + 1) mod table size [OK]

Common Mistakes:

• Adding 2 instead of 1 for next index
• Multiplying index instead of adding
• Not using modulo for wrap-around

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. [None, None, 7, 12, 17]

B. [7, 12, 17, None, None]

C. [None, 7, 12, 17, None]

D. [7, None, 12, 17, None]

Solution

1. Step 1: Calculate initial positions

   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 index 3. Insert 12 at index 3. For 17, indices 2 and 3 are taken, so insert at index 4.

3. Final Answer:

   [None, None, 7, 12, 17] -> Option A

4. Quick Check:

   Keys placed at 2, 3, 4 with linear probing [OK]

Hint: Check hash index, then next free slot sequentially [OK]

Common Mistakes:

• Placing all keys at same index ignoring collisions
• Not probing to next slot after collision
• Miscomputing hash values

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

def insert(table, key):
    m = len(table)
    index = key % m
    while table[index] is not None:
        index = (index + 1) % m
    table[index] = key

hash_table = [None]*5
insert(hash_table, 5)
insert(hash_table, 10)
insert(hash_table, 15)
insert(hash_table, 20)
insert(hash_table, 25)
insert(hash_table, 30)

medium

A. Infinite loop if table is full because no stop condition in while loop.

B. No bug; code works correctly for all insertions.

C. Wrong hash function used; should be key // m.

D. Table size should be dynamic, not fixed.

Solution

1. Step 1: Analyze insertion loop

   The while loop continues probing until it finds an empty slot but has no check for full table.

2. Step 2: Identify problem when table is full

   If all slots are occupied, the loop cycles endlessly, causing infinite loop.

3. Final Answer:

   Infinite loop if table is full because no stop condition in while loop. -> Option A

4. Quick Check:

   Full table causes infinite loop without exit [OK]

Hint: Check for full table to avoid infinite loop [OK]

Common Mistakes:

• Assuming hash function is wrong
• Ignoring infinite loop possibility
• Thinking dynamic size fixes this bug

5. You have a hash table of size 7 using linear probing. Insert keys [14, 21, 28, 35, 42] using hash function h(k) = k % 7. After all insertions, what is the index of key 42 in the table?

hard

A. 5

B. 0

C. 6

D. 4

Solution

1. Step 1: Calculate initial hash indices

   h(14)=0, h(21)=0, h(28)=0, h(35)=0, h(42)=0 (all keys hash to index 0).

2. Step 2: Insert keys using linear probing

   Insert 14 at 0. For 21, index 0 taken, try 1. Insert 21 at 1. For 28, indices 0 and 1 taken, insert at 2. For 35, insert at 3. For 42, insert at 4.

3. Step 3: Find index of key 42

   Key 42 is at index 4.

4. Final Answer:

   4 -> Option D

5. Quick Check:

   42 placed at first free slot after collisions [OK]

Hint: All keys collide; count steps from initial index [OK]

Common Mistakes:

• Assuming key stays at initial hash index
• Not probing sequentially after collisions
• Miscounting index after multiple collisions