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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Go

func hash(key int, size int) int {
    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 % causes index out of range errors.

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 using linear probing.

DSA Go

func linearProbe(index int, size int) int {
    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 - causes backward movement which is incorrect.

Adding 1 moves to the next slot in linear probing.

3fill in blank

hard

Fix the error in the insertion loop to avoid infinite loops.

DSA Go

for i := 0; i < size; i++ {
    idx := (hash(key, size) + i) [1] size
    if table[idx] == nil {
        table[idx] = &key
        break
    }
}

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

A*

B+

C-

D%

Attempts:

3 left

💡 Hint

Common Mistakes

Using + or - without modulo causes index overflow.

The modulo operator (%) ensures the index wraps around the table size to avoid out-of-range errors and infinite loops.

4fill in blank

hard

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

DSA Go

func search(key int, table []*int, size int) int {
    start := hash(key, size)
    for i := 0; i < size; i++ {
        idx := (start [1] i) [2] size
        if table[idx] == nil {
            return -1
        }
        if *table[idx] == key {
            return idx
        }
    }
    return -1
}

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

A+

B-

C%

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using - or * instead of + and % causes wrong index calculation.

We add i to the start index and use modulo (%) to wrap around the table during linear probing search.

5fill in blank

hard

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

DSA Go

func delete(key int, table []*int, size int) bool {
    start := hash(key, size)
    for i := 0; i < size; i++ {
        idx := (start [1] i) [2] size
        if table[idx] == nil {
            return false
        }
        if *table[idx] == key {
            table[idx] = [3]
            return true
        }
    }
    return false
}

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

A+

B%

Cnil

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of nil leaves old key in table.

We add i to start and use modulo (%) to wrap index. Setting table[idx] to nil deletes the key.

Practice

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

easy

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

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

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

D. Replace the existing element in the slot with the new one.

Solution

1. Step 1: Understand collision handling

   When two keys hash to the same slot, a collision occurs and must be resolved.

2. Step 2: Explain linear probing method

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

3. Final Answer:

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

4. Quick Check:

   Linear probing = sequential slot check [OK]

Hint: Linear probing checks next slots one by one [OK]

Common Mistakes:

• Confusing linear probing with chaining
• Thinking linear probing replaces existing elements
• Assuming table size doubles on collision

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 current index is i?

easy

A. (i + 1) % m

B. (i + 3) % m

C. (i - 1) % m

D. (i * 2) % m

Solution

1. Step 1: Understand linear probing index calculation

   Linear probing moves to the next slot by adding 1 to the current index.

2. Step 2: Apply modulo to wrap around

   To stay within table bounds, use modulo with table size: (i + 1) % m.

3. Final Answer:

   (i + 1) % m -> Option A

4. Quick Check:

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

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

Common Mistakes:

• Using multiplication instead of addition
• Not using modulo causing out-of-bound errors
• Subtracting instead of adding for next slot

3. Consider a hash table of size 5 with linear probing. Insert keys 7, 12, 17 in that order using hash function h(k) = k % 5. What is the final state of the table?

medium

A. [7, 12, 17, null, null]

B. [7, null, null, 12, 17]

C. [7, 12, null, 17, null]

D. [null, null, 7, 12, 17]

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 check index 3 (empty), insert 12. For 17, indexes 2 and 3 taken, check index 4 (empty), insert 17.

3. Final Answer:

   [null, null, 7, 12, 17] -> Option D

4. Quick Check:

   All keys placed sequentially after collision [OK]

Hint: Check hash, then next slots if occupied [OK]

Common Mistakes:

• Placing keys at wrong indices ignoring collisions
• Not applying modulo for hash function
• Assuming keys overwrite existing ones

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

func insert(table []int, key int) {
    m := len(table)
    index := key % m
    for table[index] != -1 {
        index = (index + 1) % m
    }
    table[index] = key
}

// Initial table: [-1, -1, -1, -1, -1]
// Insert keys: 3, 8, 13

medium

A. The loop condition should check for zero instead of -1.

B. The code does not handle the case when the table is full, causing an infinite loop.

C. The initial table should be initialized with zeros, not -1.

D. The modulo operation should be done with key, not m.

Solution

1. Step 1: Analyze loop condition and termination

   The loop continues while the slot is occupied (not -1). If the table is full, the loop never ends.

2. Step 2: Identify missing full table check

   There is no check to stop after probing all slots, so infinite loop can occur if table is full.

3. Final Answer:

   The code does not handle the case when the table is full, causing an infinite loop. -> Option B

4. Quick Check:

   Missing full table check causes infinite loop [OK]

Hint: Always check for full table to avoid infinite loops [OK]

Common Mistakes:

• Ignoring full table condition
• Confusing initial empty slot value
• Misusing modulo operation

5. You have a hash table of size 7 using linear probing. The hash function is h(k) = k % 7. You insert keys 10, 20, 15, 7, 32 in this order. After these insertions, you want to insert key 24. Which index will key 24 be placed at?

hard

A. Index 3

B. Index 4

C. Index 5

D. Index 6

Solution

1. Step 1: Insert keys and find their positions

   h(10)=3, insert at 3. h(20)=6, insert at 6. h(15)=1, insert at 1. h(7)=0, insert at 0. h(32)=4, insert at 4.

2. Step 2: Find position for key 24

   h(24)=3, but index 3 occupied by 10. Check next index 4 (occupied by 32), then 5 (empty), so insert at 5.

3. Final Answer:

   Index 5 -> Option C

4. Quick Check:

   Linear probe from 3 to 5 for key 24 [OK]

Hint: Probe next slots until empty found [OK]

Common Mistakes:

• Stopping at first collision without probing
• Not applying modulo for wrap-around
• Misplacing keys ignoring occupied slots