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DSA C++programming~10 mins

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

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to move 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'
Acount
Bn
Clength
Dsize
Attempts:
3 left
💡 Hint
Common Mistakes
Using the length of the list instead of n.
Moving the pointer less or more than n steps.
2fill in blank
medium

Complete the code to move both pointers until fast reaches the end.

DSA C++
while (fast->next != nullptr) {
    slow = slow->next;
    fast = [1];
}
Drag options to blanks, or click blank then click option'
Afast->next
Bslow->next
Cfast->prev
Dslow->prev
Attempts:
3 left
💡 Hint
Common Mistakes
Using slow pointer instead of fast.
Using prev pointer which does not exist in singly linked list.
3fill in blank
hard

Fix the error in removing the target node by completing the pointer update.

DSA C++
slow->next = [1];
Drag options to blanks, or click blank then click option'
Aslow->next->next
Bslow->prev
Cfast->next
Dnullptr
Attempts:
3 left
💡 Hint
Common Mistakes
Setting slow->next to nullptr which breaks the list.
Using fast pointer which is not at the correct position.
4fill in blank
hard

Fill both blanks to handle the case when the head node is removed.

DSA C++
if (fast == nullptr) {
    head = [1];
    return [2];
}
Drag options to blanks, or click blank then click option'
Ahead->next
Bhead->prev
Chead
Dnullptr
Attempts:
3 left
💡 Hint
Common Mistakes
Returning old head instead of updated head.
Setting head to nullptr incorrectly.
5fill in blank
hard

Fill all three blanks to complete the function signature and return statement.

DSA C++
ListNode* [1](ListNode* [2], int [3]) {
    // function body
    return head;
}
Drag options to blanks, or click blank then click option'
AremoveNthFromEnd
Bhead
Cn
DdeleteNode
Attempts:
3 left
💡 Hint
Common Mistakes
Using wrong function name.
Swapping parameter order.

Practice

(1/5)
1. What does the problem "Remove Nth Node from End of List" ask you to do with a linked list?
easy
A. Add a new node at the Nth position from the end
B. Remove the node that is Nth from the start of the list
C. Reverse the linked list
D. Remove the node that is Nth from the end of the list

Solution

  1. Step 1: Understand the problem statement

    The problem asks to remove the node that is Nth from the end of the linked list, not from the start or to reverse or add nodes.
  2. Step 2: Clarify the operation on the list

    Removing the Nth node from the end means counting nodes from the tail backwards and deleting that node.
  3. Final Answer:

    Remove the node that is Nth from the end of the list -> Option D
  4. Quick Check:

    Remove Nth from end = Remove Nth from end [OK]
Hint: Focus on counting nodes from the list's end, not start [OK]
Common Mistakes:
  • Confusing Nth from end with Nth from start
  • Thinking the problem asks to reverse the list
  • Trying to add nodes instead of removing
2. Which of the following C++ function signatures correctly represents a function to remove the Nth node from the end of a singly linked list?
easy
A. ListNode* removeNthFromEnd(ListNode* head, int n);
B. void removeNthFromEnd(int n, ListNode* head);
C. ListNode removeNthFromEnd(ListNode head, int n);
D. ListNode* removeNthFromEnd(int n);

Solution

  1. Step 1: Check parameter order and types

    The function should take the head pointer of the list and the integer n as parameters, with head first and n second.
  2. Step 2: Check return type

    Since the head might change (if first node is removed), the function returns a pointer to the new head.
  3. Final Answer:

    ListNode* removeNthFromEnd(ListNode* head, int n); -> Option A
  4. Quick Check:

    Correct signature = ListNode* removeNthFromEnd(ListNode* head, int n); [OK]
Hint: Function returns new head pointer and takes head, n as parameters [OK]
Common Mistakes:
  • Wrong parameter order
  • Returning ListNode instead of pointer
  • Missing head parameter
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 -> 3 -> 5 -> null
B. 2 -> 3 -> 4 -> 5 -> null
C. 1 -> 3 -> 4 -> 5 -> null
D. 1 -> 2 -> 4 -> 5 -> null

Solution

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

    Counting from the end: 5 (1st), 4 (2nd). So node with value 4 is to be removed.
  2. Step 2: Remove node 4 and link previous node to next

    Node 3's next pointer now points to node 5, skipping node 4.
  3. Final Answer:

    1 -> 2 -> 3 -> 5 -> null -> Option A
  4. Quick Check:

    Remove 2nd from end = Remove node 4 [OK]
Hint: Count nodes from end to find which to remove [OK]
Common Mistakes:
  • Removing wrong node counting from start
  • Not updating links correctly
  • Confusing 2nd from end with 2nd from start
4. Given the following code snippet to remove the Nth node from the end, what is the bug?
ListNode* removeNthFromEnd(ListNode* head, int n) {
    ListNode* fast = head;
    ListNode* slow = head;
    for (int i = 0; i < n; i++) {
        fast = fast->next;
    }
    while (fast->next != nullptr) {
        fast = fast->next;
        slow = slow->next;
    }
    slow->next = slow->next->next;
    return head;
}
medium
A. Uses wrong loop condition in for loop
B. Does not handle case when n equals list length (removing head)
C. Returns wrong pointer type
D. Does not move slow pointer at all

Solution

  1. Step 1: Analyze fast pointer movement

    Fast pointer moves n steps ahead. If n equals list length, fast becomes nullptr.
  2. Step 2: Check while loop condition

    While loop uses fast->next, but if fast is nullptr, accessing fast->next causes error. Also, removing head node is not handled.
  3. Final Answer:

    Does not handle case when n equals list length (removing head) -> Option B
  4. Quick Check:

    Edge case for head removal missing [OK]
Hint: Check edge cases when fast pointer reaches null early [OK]
Common Mistakes:
  • Ignoring removal of head node
  • Accessing fast->next without null check
  • Not updating head pointer when needed
5. You want to remove the 3rd node from the end of a linked list with nodes: 10 -> 20 -> 30 -> 40 -> 50 -> 60 -> null. Which of the following steps correctly describes the process using two pointers?
hard
A. Move slow pointer 3 steps ahead, then move fast pointer until slow reaches end, then remove fast's next node.
B. Move fast pointer 3 steps ahead, then move slow pointer 3 steps ahead, then remove head node.
C. Move fast pointer 3 steps ahead, then move both pointers until fast reaches end, then remove slow's next node.
D. Move both pointers together from start without offset, then remove the node at position 3.

Solution

  1. Step 1: Initialize two pointers at head

    Start both fast and slow pointers at the head of the list.
  2. Step 2: Move fast pointer n steps ahead

    Move fast pointer 3 steps ahead to create a gap of 3 nodes between fast and slow.
  3. Step 3: Move both pointers until fast reaches end

    Move fast and slow pointers one step at a time until fast reaches the last node.
  4. Step 4: Remove the node after slow pointer

    Slow pointer now points to node before the target; remove slow->next node.
  5. Final Answer:

    Move fast pointer 3 steps ahead, then move both pointers until fast reaches end, then remove slow's next node. -> Option C
  6. Quick Check:

    Two-pointer gap method = Move fast pointer 3 steps ahead, then move both pointers until fast reaches end, then remove slow's next node. [OK]
Hint: Use two pointers with gap n to find node before target [OK]
Common Mistakes:
  • Moving slow pointer first instead of fast
  • Removing wrong node after pointers move
  • Not creating correct gap between pointers
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10 questions~5 mineasy hard