Get Length of Linked List in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the length counter to zero.

DSA C

int getLength(struct Node* head) {
    int length = [1];
    struct Node* current = head;
    while (current != NULL) {
        length++;
        current = current->next;
    }
    return length;
}

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

A0

B-1

C1

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Starting length at 1 causes off-by-one errors.

Using NULL instead of 0 for an integer variable.

The length counter should start at 0 because we haven't counted any nodes yet.

2fill in blank

medium

Complete the code to move to the next node in the linked list.

DSA C

int getLength(struct Node* head) {
    int length = 0;
    struct Node* current = head;
    while (current != NULL) {
        length++;
        current = [1];
    }
    return length;
}

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

Acurrent.next

Bcurrent.next()

Chead->next

Dcurrent->next

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.' instead of '->' for pointer access.

Using function call syntax 'next()' which is invalid here.

In C, to access the next pointer of a struct pointer, use '->'.

3fill in blank

hard

Fix the error in the while loop condition to correctly check for the end of the list.

DSA C

int getLength(struct Node* head) {
    int length = 0;
    struct Node* current = head;
    while ([1]) {
        length++;
        current = current->next;
    }
    return length;
}

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

Acurrent != NULL

Bcurrent == NULL

Chead != NULL

Dhead == NULL

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'current == NULL' causes the loop to skip all nodes.

Checking 'head' inside the loop instead of 'current'.

The loop should continue while current is not NULL to traverse all nodes.

4fill in blank

hard

Fill both blanks to create a function that returns the length of the linked list.

DSA C

int getLength(struct Node* [1]) {
    int length = 0;
    struct Node* [2] = head;
    while (current != NULL) {
        length++;
        current = current->next;
    }
    return length;
}

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

Ahead

Bcurrent

Cnode

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same name for parameter and traversal pointer.

Using undefined variable names.

The function parameter is named 'head' and the traversal pointer is 'current'.

5fill in blank

hard

Fill all three blanks to complete the function that returns the length of the linked list.

DSA C

int getLength(struct Node* [1]) {
    int [2] = 0;
    struct Node* [3] = head;
    while (current != NULL) {
        length++;
        current = current->next;
    }
    return length;
}

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

Ahead

Blength

Ccurrent

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent variable names causing errors.

Not initializing the counter to zero.

The parameter is 'head', the counter is 'length', and the traversal pointer is 'current'.

Practice

1. What does the function to get the length of a linked list typically do?

easy

A. Reverses the linked list

B. Deletes all nodes in the linked list

C. Finds the maximum value stored in the linked list

D. Counts the number of nodes by traversing from head to end

Solution

1. Step 1: Understand the linked list traversal

   The function starts at the head node and moves through each node until it reaches the end (null).

2. Step 2: Count each node visited

   Each time the function moves to the next node, it increases a counter by one.

3. Final Answer:

   Counts the number of nodes by traversing from head to end -> Option D

4. Quick Check:

   Length = number of nodes visited [OK]

Hint: Count nodes while moving from head to null [OK]

Common Mistakes:

• Confusing length with maximum value
• Trying to reverse the list instead
• Not traversing all nodes

2. Which of the following is the correct C syntax to declare a pointer to the head of a linked list node named head?

easy

A. struct Node* head;

B. struct Node &head;

C. Node head*;

D. struct Node head;

Solution

1. Step 1: Recall pointer declaration in C

   To declare a pointer to a struct Node, use struct Node* variableName;.

2. Step 2: Match with options

   struct Node* head; correctly declares head as a pointer to struct Node.

3. Final Answer:

   struct Node* head; -> Option A

4. Quick Check:

   Pointer syntax uses * after type [OK]

Hint: Pointer declaration uses * after type name [OK]

Common Mistakes:

• Forgetting the * for pointer
• Using & which is reference in C++ not C
• Incorrect order of * and variable name

3. What will be the output of the following C code snippet?

struct Node {
    int data;
    struct Node* next;
};

int getLength(struct Node* head) {
    int count = 0;
    while (head != NULL) {
        count++;
        head = head->next;
    }
    return count;
}

int main() {
    struct Node n3 = {30, NULL};
    struct Node n2 = {20, &n3};
    struct Node n1 = {10, &n2};
    printf("%d", getLength(&n1));
    return 0;
}

medium

A. 3

B. 2

C. 1

D. 0

Solution

1. Step 1: Analyze linked list nodes

   There are three nodes: n1 -> n2 -> n3 -> NULL.

2. Step 2: Count nodes in getLength function

   The function counts each node until it reaches NULL, so count = 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   3 nodes counted from head to end [OK]

Hint: Count nodes until next is NULL [OK]

Common Mistakes:

• Counting only head node
• Off-by-one error in counting
• Confusing pointer addresses with count

4. Find the error in this C function to get the length of a linked list:

int getLength(struct Node* head) {
    int count = 0;
    while (head->next != NULL) {
        count++;
        head = head->next;
    }
    return count;
}

medium

A. It causes a segmentation fault immediately

B. It misses counting the last node

C. It counts nodes twice

D. It returns the data of the last node

Solution

1. Step 1: Understand the loop condition

   The loop runs while head->next != NULL, so it stops before the last node.

2. Step 2: Check counting logic

   Because it stops early, the last node is not counted, so count is one less than actual length.

3. Final Answer:

   It misses counting the last node -> Option B

4. Quick Check:

   Loop condition excludes last node [OK]

Hint: Use while(head != NULL) to count all nodes [OK]

Common Mistakes:

• Using head->next in loop condition
• Not handling empty list (NULL head)
• Incrementing count after loop ends

5. You have a linked list where some nodes may have next pointers forming a cycle (loop). How would you modify the length function to avoid infinite counting?

hard

A. Stop counting after 10 nodes to avoid infinite loop

B. Count nodes normally without any change

C. Use two pointers moving at different speeds to detect a cycle before counting

D. Set all next pointers to NULL before counting

Solution

1. Step 1: Detect cycle using two pointers

   Use Floyd's cycle detection: one pointer moves one step, another moves two steps to find a loop.

2. Step 2: If cycle detected, handle counting carefully

   Once a cycle is found, count nodes up to the cycle start to avoid infinite counting.

3. Final Answer:

   Use two pointers moving at different speeds to detect a cycle before counting -> Option C

4. Quick Check:

   Cycle detection prevents infinite loops [OK]

Hint: Detect cycle first with slow and fast pointers [OK]

Common Mistakes:

• Ignoring cycle causes infinite loop
• Stopping count arbitrarily without detection
• Modifying list structure destructively