Remove Nth Node from End of List in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with given value.

DSA C

struct ListNode* newNode = (struct ListNode*)malloc(sizeof(struct ListNode));
newNode->val = [1];
newNode->next = NULL;

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

Ax

Bvalue

Cn

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using an undefined variable name like 'val' or 'value'.

Confusing the node pointer with the value variable.

The variable 'x' holds the value to assign to the new node's val field.

2fill in blank

medium

Complete the code to advance the fast pointer n steps ahead.

DSA C

for (int i = 0; i < [1]; i++) {
    fast = fast->next;
}

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

Alength

Bcount

Cn

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable like length or count which are unrelated here.

Off-by-one errors in the loop condition.

We need to move the fast pointer n steps ahead, so the loop runs n times.

3fill in blank

hard

Fix the error in the condition to check if fast pointer reached the end.

DSA C

if (fast == [1]) {
    return head->next;
}

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

Afast->next

BNULL

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Checking against head or tail which are not correct here.

Using fast->next which can cause a null pointer dereference.

When fast reaches NULL, it means we need to remove the head node.

4fill in blank

hard

Fill both blanks to move fast and slow pointers until fast reaches the end.

DSA C

while (fast->next != [1]) {
    fast = fast->next;
    slow = [2];
}

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

ANULL

Bslow->next

Chead

Dfast

Attempts:

3 left

💡 Hint

Common Mistakes

Using head or fast in place of slow->next.

Checking fast->next against head instead of NULL.

We move fast until fast->next is NULL, and move slow one step forward each time.

5fill in blank

hard

Fill all three blanks to remove the nth node from end by adjusting pointers.

DSA C

struct ListNode* toDelete = slow->[1];
slow->[2] = toDelete->[3];
free(toDelete);

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

Anext

Bprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev pointers which do not exist in singly linked list.

Not freeing the deleted node causing memory leaks.

We set toDelete as slow->next, then link slow->next to toDelete->next, and free toDelete.

Practice

1. What does the problem "Remove Nth Node from End of List" ask you to do with a linked list?

easy

A. Remove the node that is Nth from the end of the list

B. Remove the node that is Nth from the start of the list

C. Add a new node at the Nth position from the end

D. Reverse the linked list up to the Nth node from the end

Solution

1. Step 1: Understand the problem statement

   The problem requires removing the node that is exactly N positions from the end of the linked list.

2. Step 2: Clarify the difference from other options

   Options A, C, and D describe different operations: removing from start, adding nodes, or reversing, which are not asked here.

3. Final Answer:

   Remove the node that is Nth from the end of the list -> Option A

4. Quick Check:

   Remove Nth from end = Remove the node that is Nth from the end of the list [OK]

Hint: Focus on 'Nth from end' means counting backward [OK]

Common Mistakes:

• Confusing Nth from start with Nth from end
• Thinking the problem asks to add nodes
• Assuming the list needs reversing

2. Which of the following is the correct way to declare a pointer to the head of a singly linked list in C?

easy

A. struct Node head*;

B. Node head*;

C. struct Node* head;

D. Node* head;

Solution

1. Step 1: Recall C struct pointer syntax

   In C, to declare a pointer to a struct Node, use 'struct Node* head;'.

2. Step 2: Identify incorrect syntax

   Options A, B, and D misuse the order or omit 'struct' keyword which is required unless typedef is used.

3. Final Answer:

   struct Node* head; -> Option C

4. Quick Check:

   Pointer to struct Node = struct Node* head; [OK]

Hint: Remember 'struct' keyword before struct name in C [OK]

Common Mistakes:

• Omitting 'struct' keyword in declaration
• Placing '*' incorrectly after variable name
• Using typedef name without defining it

3. What will be the output linked list after removing the 2nd node from the end in the list: 1 -> 2 -> 3 -> 4 -> 5 -> null?

medium

A. 1 -> 2 -> 4 -> 5 -> null

B. 2 -> 3 -> 4 -> 5 -> null

C. 1 -> 3 -> 4 -> 5 -> null

D. 1 -> 2 -> 3 -> 5 -> null

Solution

1. Step 1: Identify the 2nd node from the end

   The list is 1->2->3->4->5. Counting from end: 5 (1st), 4 (2nd). So node with value 4 is to be removed.

2. Step 2: Remove node with value 4 and link previous to next

   After removal, the list becomes 1->2->3->5->null.

3. Final Answer:

   1 -> 2 -> 3 -> 5 -> null -> Option D

4. Quick Check:

   Remove 2nd from end (4) = 1->2->3->5->null [OK]

Hint: Count nodes from end to find which to remove [OK]

Common Mistakes:

• Removing the wrong node counting from start
• Skipping the link update after removal
• Confusing 2nd from end with 2nd from start

4. Given the following code snippet to remove the Nth node from end, what is the error?

struct Node* removeNthFromEnd(struct Node* head, int n) {
    struct Node* fast = head;
    struct Node* slow = head;
    for (int i = 0; i < n; i++) {
        fast = fast->next;
    }
    while (fast->next != NULL) {
        fast = fast->next;
        slow = slow->next;
    }
    slow->next = slow->next->next;
    return head;
}

medium

A. The slow pointer is moved too many times causing segmentation fault

B. The loop does not check if fast becomes NULL before accessing fast->next

C. The function does not handle the case when n is zero

D. The function returns the wrong pointer instead of head

Solution

1. Step 1: Analyze the for loop moving fast pointer

   The loop moves fast pointer n times without checking if fast is NULL, which can cause a crash if n is larger than list length.

2. Step 2: Check while loop condition and pointer usage

   The while loop uses fast->next without null check before the loop, which is unsafe if fast is NULL.

3. Final Answer:

   The loop does not check if fast becomes NULL before accessing fast->next -> Option B

4. Quick Check:

   Missing NULL check on fast pointer causes error [OK]

Hint: Always check pointers for NULL before dereferencing [OK]

Common Mistakes:

• Not checking if fast is NULL before fast->next
• Assuming list length is always >= n
• Ignoring edge case when removing head node

5. You have a linked list: 10 -> 20 -> 30 -> 40 -> 50 -> null. You want to remove the 5th node from the end. What will be the resulting list?

hard

A. 20 -> 30 -> 40 -> 50 -> null

B. 10 -> 20 -> 30 -> 40 -> null

C. 10 -> 20 -> 30 -> 40 -> 50 -> null (no change)

D. 30 -> 40 -> 50 -> null

Solution

1. Step 1: Identify the 5th node from the end

   The list length is 5, so the 5th node from the end is the first node with value 10.

2. Step 2: Remove the head node and update head pointer

   Removing the first node results in the list: 20 -> 30 -> 40 -> 50 -> null.

3. Final Answer:

   20 -> 30 -> 40 -> 50 -> null -> Option A

4. Quick Check:

   Remove 5th from end (head) = list without first node [OK]

Hint: Removing N equals list length means remove head [OK]

Common Mistakes:

• Not updating head pointer when removing first node
• Confusing Nth from end with Nth from start
• Assuming no change if N equals list length