Huffman Encoding in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node for the Huffman tree.

DSA C

struct Node* createNode(char data, unsigned freq) {
    struct Node* temp = (struct Node*)malloc(sizeof(struct Node));
    temp->data = [1];
    temp->freq = freq;
    temp->left = temp->right = NULL;
    return temp;
}

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

Adata

Bfreq

C'\0'

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning frequency to data field.

Assigning NULL or zero character instead of actual data.

The node's data field should be set to the character passed as 'data'.

2fill in blank

medium

Complete the code to check if a node is a leaf node in the Huffman tree.

DSA C

int isLeaf(struct Node* root) {
    return (root->left == NULL && root->right [1] NULL);
}

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

A<

B>

C!=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' causing wrong leaf detection.

Using relational operators like '>' or '<' which are invalid for pointers.

A leaf node has both left and right pointers equal to NULL.

3fill in blank

hard

Fix the error in the code that inserts a node into the min heap.

DSA C

void insertMinHeap(struct MinHeap* minHeap, struct Node* node) {
    ++minHeap->size;
    int i = minHeap->size - 1;
    while (i && node->freq < minHeap->array[(i - 1) / 2]->freq) {
        minHeap->array[i] = minHeap->array[(i - 1) / 2];
        i = [1];
    }
    minHeap->array[i] = node;
}

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

Ai / 2

B(i - 1) / 2

Ci - 1

Di + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using i - 1 moves one step left, not to parent.

Using i / 2 is incorrect for zero-based indexing.

Using i + 1 moves down, not up.

To move up the heap, update i to the parent's index: (i - 1) / 2.

4fill in blank

hard

Fill both blanks to build the Huffman tree by extracting two minimum nodes and combining them.

DSA C

while (minHeap->size != 1) {
    struct Node* left = extractMin(minHeap);
    struct Node* right = extractMin(minHeap);
    struct Node* top = createNode('[1]', left->freq + right->freq);
    top->left = left;
    top->right = right;
    insertMinHeap(minHeap, top);
}

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

A'\0'

B'*'

C'$'

D' '

Attempts:

3 left

💡 Hint

Common Mistakes

Using actual characters from input instead of a special marker.

Using null character '\0' which may cause confusion.

The internal nodes use a special character like '$' to indicate they are not leaves.

5fill in blank

hard

Fill the blanks to print the Huffman codes by traversing the tree recursively.

DSA C

void printCodes(struct Node* root, int arr[], int top) {
    if (root->left) {
        arr[top] = [1];
        printCodes(root->left, arr, top + 1);
    }
    if (root->right) {
        arr[top] = [2];
        printCodes(root->right, arr, top + 1);
    }
    if (isLeaf(root)) {
        printf("%c: ", root->data);
        for (int i = 0; i < top; ++i) {
            printf("%d", arr[i]);
        }
        printf("\n");
    }
}

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

A0

B1

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 2 or -1 which are invalid for binary codes.

Swapping 0 and 1 causing incorrect codes.

Left edges are marked with 0, right edges with 1. The option '2' is a distractor and not used here.

Practice

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

easy

A. To assign equal length codes to all characters

B. To assign shorter codes to more frequent characters

C. To encrypt data for security

D. To sort characters alphabetically

Solution

1. Step 1: Understand Huffman Encoding purpose

   Huffman Encoding compresses data by giving shorter codes to characters that appear more often.

2. Step 2: Compare options with this purpose

   Only To assign shorter codes to more frequent characters matches this goal; others describe different concepts.

3. Final Answer:

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

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 compression with encryption
• Assuming sorting is the main goal

2. Which data structure is primarily used to build the Huffman tree?

easy

A. Priority Queue (Min-Heap)

B. Stack

C. Queue

D. Hash Table

Solution

1. Step 1: Identify data structure for selecting smallest frequencies

   Huffman tree construction repeatedly picks two smallest frequency nodes.

2. Step 2: Match with data structure that supports this

   Priority Queue (Min-Heap) efficiently provides smallest elements first.

3. Final Answer:

   Priority Queue (Min-Heap) -> Option A

4. Quick Check:

   Smallest freq nodes from Min-Heap [OK]

Hint: Huffman uses Min-Heap to pick smallest frequencies fast [OK]

Common Mistakes:

• Choosing Stack or Queue which don't sort by frequency
• Confusing Hash Table with frequency sorting
• Ignoring the need for smallest frequency extraction

3. What is the output Huffman code for characters with frequencies: A=5, B=9, C=12, D=13, E=16, F=45?

Assuming standard Huffman tree construction, what is the code for character 'F'?

medium

A. 0

B. 10

C. 110

D. 111

Solution

1. Step 1: Build Huffman tree with given frequencies

   Combine smallest pairs: (A=5,B=9)=14, (C=12,D=13)=25, then combine 14 and 16 (E) = 30, then 25 and 30 = 55, finally combine 45 (F) and 55 = 100.

2. Step 2: Assign codes from root; 'F' is combined last and placed on left branch

   Since 'F' has highest frequency, it gets the shortest code '0'.

3. Final Answer:

   0 -> Option A

4. Quick Check:

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

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

Common Mistakes:

• Assigning longer codes to highest frequency character
• Mixing up left/right branch codes
• Not building tree step-by-step

4. Consider this C code snippet for building a Huffman tree:

struct Node {
  char data;
  int freq;
  struct Node *left, *right;
};

// Function to create a new node
struct Node* newNode(char data, int freq) {
  struct Node* temp = malloc(sizeof(struct Node));
  temp->data = data;
  temp->freq = freq;
  temp->left = temp->right = NULL;
  return temp;
}

// Code to combine two nodes
struct Node* combineNodes(struct Node* left, struct Node* right) {
  struct Node* parent = newNode('$', left->freq + right->freq);
  parent->left = left;
  parent->right = right;
  return parent;
}

// Usage:
struct Node* n1 = newNode('A', 5);
struct Node* n2 = newNode('B', 9);
struct Node* parent = combineNodes(n1, n2);

printf("Parent freq: %d\n", parent->freq);

What is the output of this code?

medium

A. Parent freq: 5

B. Parent freq: 9

C. Compilation error

D. Parent freq: 14

Solution

1. Step 1: Understand node creation and combination

   n1 has freq 5, n2 has freq 9; combineNodes sums frequencies: 5 + 9 = 14.

2. Step 2: Print parent frequency

   parent->freq is 14, so printf outputs "Parent freq: 14".

3. Final Answer:

   Parent freq: 14 -> Option D

4. Quick Check:

   5 + 9 = 14 frequency sum [OK]

Hint: Sum frequencies when combining nodes for parent freq [OK]

Common Mistakes:

• Printing child node frequency instead of parent
• Forgetting to sum frequencies
• Assuming compilation error due to struct usage

5. You have characters with frequencies: X=3, Y=3, Z=3, W=3. Using Huffman Encoding, what is the minimum total number of bits needed to encode the string "XYZWXYZW" (8 characters)?

hard

A. 12 bits

B. 24 bits

C. 16 bits

D. 8 bits

Solution

1. Step 1: Build Huffman tree for equal frequencies

   All characters have equal frequency (3), so Huffman codes will have equal length, typically 2 bits each (since 4 chars need 2 bits minimum).

2. Step 2: Calculate total bits for 8 characters

   Each character uses 2 bits, total bits = 8 * 2 = 16 bits.

3. Step 3: Re-examine options and problem

   Options include 16 and 24 bits. Since each char is 2 bits, total is 16 bits, but 24 bits is 24 bits. The string length is 8, so 8*2=16 bits is correct.

4. Final Answer:

   16 bits -> Option C

5. Quick Check:

   4 chars need 2 bits each, 8 chars total = 16 bits [OK]

Hint: Equal freq chars get equal length codes; multiply length by count [OK]

Common Mistakes:

• Assuming 1 bit per char incorrectly
• Multiplying frequency instead of code length
• Choosing 24 bits thinking 3 bits per char