Huffman Encoding in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a frequency map from the input string.

DSA Typescript

const frequencyMap: Record<string, number> = {};
for (const char of input) {
  frequencyMap[char] = (frequencyMap[char] || 0) [1] 1;
}

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition.

Using multiplication or division which doesn't make sense here.

We add 1 to the current count for each character to build the frequency map.

2fill in blank

medium

Complete the code to sort the nodes by frequency in ascending order.

DSA Typescript

nodes.sort((a, b) => a.frequency [1] b.frequency);

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

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition which does not sort correctly.

Using multiplication or division which are invalid here.

Subtracting frequencies sorts the array in ascending order.

3fill in blank

hard

Fix the error in the code that merges two nodes into a new parent node.

DSA Typescript

const mergedNode = {
  character: null,
  frequency: left.frequency [1] right.frequency,
  left: left,
  right: right
};

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

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction which can cause negative frequencies.

Using multiplication or division which are incorrect here.

The frequency of the merged node is the sum of the two child nodes' frequencies.

4fill in blank

hard

Fill both blanks to assign codes recursively to the Huffman tree nodes.

DSA Typescript

function assignCodes(node: Node | null, code: string) {
  if (!node) return;
  if (node.character !== null) {
    codes[node.character] = code;
  }
  assignCodes(node.[1], code + '0');
  assignCodes(node.[2], code + '1');
}

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

Aleft

Bright

Cparent

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using parent or root which are not children nodes.

Swapping left and right which reverses codes.

We traverse left child with '0' and right child with '1' to assign codes.

5fill in blank

hard

Fill all three blanks to build the Huffman tree by merging nodes until one remains.

DSA Typescript

while (nodes.length > 1) {
  const left = nodes.shift();
  const right = nodes.shift();
  const mergedNode = {
    character: null,
    frequency: left.frequency [1] right.frequency,
    left: left,
    right: right
  };
  nodes.push(mergedNode);
  nodes.sort((a, b) => a.frequency [2] b.frequency);
}
const root = nodes.[3]();

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

A+

B-

Cpop

Dshift

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition for frequency sum.

Using addition instead of subtraction in sort comparator.

Using shift instead of pop to get the root node.

We sum frequencies to merge nodes, sort ascending by subtracting frequencies, and finally pop the root node.

Practice

1. What is the main purpose of Huffman Encoding in data compression?

easy

A. To sort characters alphabetically

B. To assign equal length codes to all characters

C. To encrypt data for security

D. To assign shorter codes to more frequent characters

Solution

1. Step 1: Understand Huffman Encoding goal

   Huffman Encoding aims to reduce the total bits used by giving shorter codes to frequent characters.

2. Step 2: Compare options with this goal

   Only To assign shorter codes to more frequent characters matches this goal. Others describe different or incorrect purposes.

3. Final Answer:

   To assign shorter codes to more frequent characters -> Option D

4. Quick Check:

   Huffman Encoding = shorter codes for frequent chars [OK]

Hint: Remember: frequent chars get shorter codes in Huffman [OK]

Common Mistakes:

• Thinking Huffman assigns equal length codes
• Confusing Huffman with encryption
• Assuming it sorts characters alphabetically

2. Which TypeScript data structure is best suited to build the priority queue for Huffman Encoding?

easy

A. Array with manual sorting

B. Min-Heap

C. Set

D. Map

Solution

1. Step 1: Identify priority queue needs

   Huffman Encoding requires repeatedly extracting the smallest frequency node efficiently.

2. Step 2: Choose data structure for efficient min extraction

   Min-Heap allows quick access and removal of the smallest element, ideal for this use.

3. Final Answer:

   Min-Heap -> Option B

4. Quick Check:

   Priority queue = Min-Heap for Huffman [OK]

Hint: Use Min-Heap for efficient smallest element extraction [OK]

Common Mistakes:

• Using Set which has no order
• Using Map which is key-value, not priority queue
• Using Array without sorting is inefficient

3. What is the output Huffman code for characters {a:5, b:9, c:12, d:13, e:16, f:45} after building the Huffman tree?

medium

A. { 'f': '1', 'c': '00', 'd': '01', 'a': '100', 'b': '101', 'e': '110' }

B. { 'a': '0', 'b': '10', 'c': '110', 'd': '1110', 'e': '1111', 'f': '101' }

C. { 'f': '0', 'c': '100', 'd': '101', 'a': '1100', 'b': '1101', 'e': '111' }

D. { 'a': '111', 'b': '110', 'c': '10', 'd': '0', 'e': '101', 'f': '100' }

Solution

1. Step 1: Build Huffman tree from frequencies

   Combine lowest frequencies stepwise: (a=5,b=9)->14, (c=12,d=13)->25, then combine 14 and 16 (e) and so on, finally f=45 is highest frequency with shortest code.

2. Step 2: Assign codes from tree paths

   Left edges as '0', right edges as '1' produce codes: f='0', c='100', d='101', a='1100', b='1101', e='111'.

3. Final Answer:

   { 'f': '0', 'c': '100', 'd': '101', 'a': '1100', 'b': '1101', 'e': '111' } -> Option C

4. Quick Check:

   Highest freq 'f' gets shortest code '0' [OK]

Hint: Highest frequency gets shortest code '0' in Huffman [OK]

Common Mistakes:

• Assigning codes without building tree
• Mixing left/right edge bits
• Ignoring frequency order

4. Given this TypeScript snippet for Huffman tree node insertion, what is the bug?

class Node {
  constructor(public char: string | null, public freq: number, public left: Node | null = null, public right: Node | null = null) {}
}

function insertNode(queue: Node[], node: Node) {
  queue.push(node);
  queue.sort((a, b) => a.freq - b.freq);
}

// Usage
const queue: Node[] = [];
insertNode(queue, new Node('a', 5));
insertNode(queue, new Node('b', 9));
insertNode(queue, new Node(null, 14));

medium

A. Sorting after push is inefficient; use a Min-Heap instead

B. Node constructor missing type annotations

C. Function insertNode should remove duplicates

D. queue.sort comparator should be reversed

Solution

1. Step 1: Analyze insertNode function

   It pushes a node then sorts the entire queue array by frequency each time.

2. Step 2: Identify inefficiency

   Sorting the whole array on every insertion is inefficient for large data; a Min-Heap is better for priority queue operations.

3. Final Answer:

   Sorting after push is inefficient; use a Min-Heap instead -> Option A

4. Quick Check:

   Priority queue needs efficient insert and extract [OK]

Hint: Use Min-Heap, not array sort, for priority queue [OK]

Common Mistakes:

• Thinking sorting order is wrong
• Missing inefficiency of sorting whole array
• Confusing type annotations as bug

5. You have a string "aaabbc" and want to encode it using Huffman Encoding. Which of the following is the correct Huffman code assignment and encoded bit string?

hard

A. Codes: {a:'0', b:'10', c:'11'}, Encoded: '000101011'

B. Codes: {a:'10', b:'0', c:'11'}, Encoded: '101010011'

C. Codes: {a:'0', b:'11', c:'10'}, Encoded: '00011110'

D. Codes: {a:'1', b:'0', c:'10'}, Encoded: '1110010'

Solution

1. Step 1: Calculate frequencies

   String "aaabbc" has frequencies: a=3, b=2, c=1.

2. Step 2: Build Huffman tree and assign codes

   Lowest freq c=1 and b=2 combine first, then with a=3. Assign codes: a='0', b='10', c='11'.

3. Step 3: Encode string using codes

   Encoding "aaabbc" = a(3 times) '0 0 0', b(2 times) '10 10', c(1 time) '11' -> '000101011'.

4. Final Answer:

   Codes: {a:'0', b:'10', c:'11'}, Encoded: '000101011' -> Option A

5. Quick Check:

   Freq order a>b>c matches codes and encoding [OK]

Hint: Encode by replacing chars with shortest codes from tree [OK]

Common Mistakes:

• Mixing code assignments
• Encoding with wrong code bits
• Ignoring frequency order in code assignment