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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

function hash(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 - causes index out of range errors.

The modulo operator (%) ensures the hash index stays within the table size.

2fill in blank

medium

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

DSA Javascript

function nextIndex(currentIndex, size) {
  return (currentIndex [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 here.

Adding 1 moves to the next slot in linear probing.

3fill in blank

hard

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

DSA Javascript

while (table[index] !== null && table[index].key !== key) {
  index = (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 infinite loops or wrong slot checking.

Incrementing index by 1 moves to the next slot for linear probing.

4fill in blank

hard

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

DSA Javascript

function search(table, key, size) {
  let index = key [1] size;
  while (table[index] !== null && table[index].key !== key) {
    index = (index [2] 1) % size;
  }
  return table[index];
}

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

A%

B+

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using - instead of + causes backward search.

The initial index is found using modulo (%). Then we add 1 to move to the next slot during probing.

5fill in blank

hard

Fill all three blanks to complete the insert function with linear probing collision handling.

DSA Javascript

function insert(table, key, value, size) {
  let index = key [1] size;
  while (table[index] !== null && table[index].key !== key) {
    index = (index [2] 1) % size;
  }
  table[index] = { key: key, value: [3] };
}

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

A%

B+

Cvalue

Dkey

Attempts:

3 left

💡 Hint

Common Mistakes

Using key instead of value stores wrong data.

We use modulo (%) to find the start index, add 1 to probe next slots, and store the given value in the table.

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 values.

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

D. Replace the existing value in the slot with the new value.

Solution

1. Step 1: Understand collision handling methods

   Linear probing handles collisions by searching the next available slot in a sequence.

2. Step 2: Identify linear probing behavior

   When a collision happens, linear probing checks the next slot one by one until it finds an empty slot.

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 means try next slot one by one [OK]

Common Mistakes:

• Confusing linear probing with chaining
• Thinking collision means resizing immediately
• Replacing existing values instead of 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 current index is i?

easy

A. (i * 2) % m

B. (i - 1) % m

C. (i + 3) % m

D. (i + 1) % m

Solution

1. Step 1: Recall linear probing index update

   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 size, use modulo operation with table size m.

3. Final Answer:

   (i + 1) % m -> Option D

4. Quick Check:

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

Hint: Next index in linear probing is (i + 1) mod table size [OK]

Common Mistakes:

• Using multiplication instead of addition
• Not using modulo to wrap index
• Subtracting instead of adding

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 (index: value)?

medium

A. 0: null, 1: null, 2: 7, 3: 12, 4: 17

B. 0: null, 1: 7, 2: null, 3: 12, 4: 17

C. 0: 7, 1: 12, 2: 17, 3: null, 4: null

D. 0: null, 1: 7, 2: 12, 3: 17, 4: null

Solution

1. Step 1: Calculate initial positions using hash function

   7 % 5 = 2, 12 % 5 = 2, 17 % 5 = 2. All keys initially map to index 2.

2. Step 2: Insert keys using linear probing

   Insert 7 at index 2 (empty). Insert 12: index 2 occupied, try 3 (empty), insert at 3. Insert 17: index 2 occupied, 3 occupied, try 4 (empty), insert at 4.

3. Final Answer:

   0: null, 1: null, 2: 7, 3: 12, 4: 17 -> Option A

4. Quick Check:

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

Hint: All keys hash to 2, probe next slots for empty [OK]

Common Mistakes:

• Placing all keys at index 2 ignoring collisions
• Skipping slots incorrectly during probing
• Miscomputing hash values

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

const size = 7;
const table = new Array(size).fill(null);
function hash(key) {
  return key % size;
}
function insert(key) {
  let index = hash(key);
  while (table[index] !== null) {
    index = (index + 1) % size;
  }
  table[index] = key;
}
insert(10);
insert(17);
insert(24);
console.log(table);

medium

A. The hash function should use key % (size - 1) instead of key % size.

B. The table array should be initialized with zeros instead of null.

C. The insert function does not handle the case when the table is full, causing an infinite loop.

D. The index update in the while loop should subtract 1 instead of adding 1.

Solution

1. Step 1: Analyze insert function loop

   The while loop probes next slots until it finds an empty slot (null).

2. Step 2: Identify missing full table check

   If the table is full, the loop cycles endlessly without exit condition, causing infinite loop.

3. Final Answer:

   The insert function does not handle the case when the table is full, causing an infinite loop. -> Option C

4. Quick Check:

   Missing full table check causes infinite loop [OK]

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

Common Mistakes:

• Thinking modulo should be size-1
• Ignoring infinite loop possibility
• Confusing null with zero initialization

5. You have a hash table of size 10 using linear probing. The table currently stores keys at indices: 2, 3, 4, 5. You want to insert key 22. The hash function is h(k) = k % 10. Which index will key 22 be inserted into?

hard

A. 2

B. 6

C. 7

D. 5

Solution

1. Step 1: Calculate initial index for key 22

   22 % 10 = 2, so initial index is 2.

2. Step 2: Apply linear probing to find empty slot

   Indices 2, 3, 4, 5 are occupied. Next index is 6, which is free.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Start at 2, probe until free slot at 6 [OK]

Hint: Probe forward from hash index until empty slot found [OK]

Common Mistakes:

• Inserting at first hashed index ignoring occupancy
• Stopping probing too early
• Choosing wrong next index