Collision Handling Using Chaining in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new empty bucket (linked list) at the given index.

DSA Javascript

if (!this.buckets[index]) {
  this.buckets[index] = [1];
}

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

Anew LinkedList()

B{}

Cnull

D[]

Attempts:

3 left

💡 Hint

Common Mistakes

Using an array [] instead of a linked list.

Setting the bucket to null or an empty object {}.

We use a linked list to handle collisions by chaining, so we create a new LinkedList at the bucket index.

2fill in blank

medium

Complete the code to add a key-value pair to the linked list at the bucket index.

DSA Javascript

this.buckets[index].[1]({ key, value });

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

Aremove

Bappend

Cfind

Dupdate

Attempts:

3 left

💡 Hint

Common Mistakes

Using find or update which do not add new nodes.

Using remove which deletes nodes.

We append the new key-value pair to the linked list to handle collisions by chaining.

3fill in blank

hard

Fix the error in the code to find a node with the matching key in the linked list.

DSA Javascript

const node = this.buckets[index].find(node => node.key [1] key);

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

A==

B=

C!=

D===

Attempts:

3 left

💡 Hint

Common Mistakes

Using single = which assigns instead of compares.

Using == which is loose equality and can cause bugs.

Use strict equality === to correctly compare keys in JavaScript.

4fill in blank

hard

Fill both blanks to update the value if the key exists, or append a new node if it does not.

DSA Javascript

if (node) {
  node.value = [1];
} else {
  this.buckets[index].[2]({ key, value });
}

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

Avalue

Bappend

Ckey

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning key instead of value in the update.

Using remove instead of append to add new nodes.

If the node exists, update its value. Otherwise, append a new node with the key and value.

5fill in blank

hard

Fill all three blanks to remove a key-value pair from the linked list at the bucket index.

DSA Javascript

const removed = this.buckets[index].[1](node => node.key [2] [3]);

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

Aremove

B===

Ckey

Dfind

Attempts:

3 left

💡 Hint

Common Mistakes

Using find instead of remove to delete nodes.

Using loose equality or assignment operators.

Use remove method with a callback that checks if node.key strictly equals the key to remove.

Practice

1. What is the main idea behind collision handling using chaining in hash tables?

easy

A. Resize the hash table immediately after every insertion

B. Replace the existing element with the new one when a collision happens

C. Store multiple elements in a linked list at the same index when collisions occur

D. Use a secondary hash function to find a new index

Solution

1. Step 1: Understand collision in hash tables

   When two keys hash to the same index, a collision occurs.

2. Step 2: How chaining handles collisions

   Chaining stores all collided elements in a linked list at that index.

3. Final Answer:

   Store multiple elements in a linked list at the same index when collisions occur -> Option C

4. Quick Check:

   Chaining = linked list at index [OK]

Hint: Chaining means linked list at one index for collisions [OK]

Common Mistakes:

• Thinking chaining replaces old values
• Confusing chaining with open addressing
• Assuming resizing happens on every collision

2. Which of the following is the correct way to initialize a hash table with chaining in JavaScript?

easy

A. const table = new Array(10).fill([]);

B. const table = Array.from({ length: 10 }, () => []);

C. const table = new Array(10);

D. const table = Array(10).fill(null);

Solution

1. Step 1: Understand array initialization with unique lists

   Using fill([]) fills all indexes with the same array reference.

2. Step 2: Use Array.from with a function to create unique arrays

   Array.from({ length: 10 }, () => []) creates a new empty array at each index.

3. Final Answer:

   const table = Array.from({ length: 10 }, () => []); -> Option B

4. Quick Check:

   Unique arrays per index = const table = Array.from({ length: 10 }, () => []); [OK]

Hint: Use Array.from with function for unique arrays [OK]

Common Mistakes:

• Using fill([]) causes all indexes to share one array
• new Array(10) leaves slots undefined
• Filling with null does not create lists for chaining

3. Consider the following JavaScript code for inserting keys into a hash table with chaining:

const table = Array.from({ length: 5 }, () => []);
function hash(key) { return key % 5; }

// Insert keys
[7, 12, 17].forEach(key => {
  const index = hash(key);
  table[index].push(key);
});

console.log(table);

What is the output printed?

medium

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

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

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

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

Solution

1. Step 1: Calculate hash index for each key

   7 % 5 = 2, 12 % 5 = 2, 17 % 5 = 2.

2. Step 2: Insert keys at index 2

   All keys go into table[2], so table[2] = [7, 12, 17].

3. Final Answer:

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

4. Quick Check:

   All keys at index 2 = [[], [], [7, 12, 17], [], []] [OK]

Hint: Calculate key % size to find index [OK]

Common Mistakes:

• Miscomputing modulo results
• Inserting keys at wrong indexes
• Confusing array positions in output

4. The following code attempts to insert keys into a hash table with chaining but has a bug:

const table = new Array(3).fill([]);
function hash(key) { return key % 3; }

function insert(key) {
  const index = hash(key);
  table[index].push(key);
}

insert(1);
insert(4);
insert(7);
console.log(table);

What is the bug and how to fix it?

medium

A. Bug: All indexes share the same array; Fix: Use Array.from with () => []

B. Bug: hash function is incorrect; Fix: Use key % 10 instead

C. Bug: push method is wrong; Fix: Use unshift instead

D. Bug: table size is too small; Fix: Increase size to 10

Solution

1. Step 1: Identify the bug with fill([])

   Using fill([]) assigns the same array reference to all indexes.

2. Step 2: Fix by creating unique arrays per index

   Use Array.from({ length: 3 }, () => []) to create separate arrays.

3. Final Answer:

   Bug: All indexes share the same array; Fix: Use Array.from with () => [] -> Option A

4. Quick Check:

   fill([]) shares arrays, Array.from creates unique ones [OK]

Hint: Avoid fill([]); use Array.from with function [OK]

Common Mistakes:

• Changing hash function unnecessarily
• Using wrong array methods
• Ignoring shared reference problem

5. You have a hash table of size 4 using chaining. You insert keys [3, 7, 11, 15, 19]. The hash function is key % 4. After all insertions, which index has the longest chain and what is its content?

medium

A. Index 3: [3, 7, 11, 15, 19]

B. Index 1: [3, 7, 11, 15, 19]

C. Index 3: [3, 7, 11, 15]

D. Index 3: [3, 7, 11]

Solution

1. Step 1: Compute hash index for each key

   3 % 4 = 3, 7 % 4 = 3, 11 % 4 = 3, 15 % 4 = 3, 19 % 4 = 3.

2. Step 2: Insert all keys at index 3

   All keys go into table[3], so chain is [3, 7, 11, 15, 19].

3. Final Answer:

   Index 3: [3, 7, 11, 15, 19] -> Option A

4. Quick Check:

   All keys mod 4 = 3, chain length 5 [OK]

Hint: All keys mod 4 equal 3, so chain at index 3 [OK]

Common Mistakes:

• Miscomputing modulo for some keys
• Assuming keys spread across indexes
• Forgetting all keys collide at same index