LeetCode Remove Nth Node From End of List
Problem statement
Given the head of a linked list, remove the nth node from the end of the list and return its head.
Example 1:
Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]
Example 2:
Input: head = [1], n = 1
Output: []
Example 3:
Input: head = [1,2], n = 1
Output: [1]
Constraints:
 The number of nodes in the list is sz.
 1 <= sz <= 30
 0 <= Node.val <= 100
 1 <= n <= sz
Explanation
Single pointer
One of the approaches to solve this problem is use single pointer by following the below steps
 Calculating the length of the linked list
 Subtract n from the length
 Start from the head and iterate to above (lengthn)th node.
A C++ snippet for the above solution is as below:
ListNode* first = head;
while (first != null) {
length++;
first = first.next;
}
length = n;
first = dummy;
while (length > 0) {
length;
first = first.next;
}
first.next = first.next.next;
// dummy next is pointing to the head of the list.
return dummy.next;
The above solution is fine, but the main concern here is the repeated iteration on the linked list.
Consider a case where the list is very huge of length 1,000,000 and we need to remove the 5th node from last. With the above approach, we are iterating over the list twice.
Two pointer
We can use two pointers and remove the node from the list in a single pass. Let's check the algorithm for this.
Algorithm
 Initialize two pointers slow and fast pointing to the head of the list.
 Loop while n > 0
 fast = fast>next
 decrement n
// if fast is nil it means the first node is supposed to be removed
 if fast == nil
 head = head>next
 return head
 Loop while fast>next != nil
 slow = slow>next
 fast = fast>next
 if slow>next != nil && slow>next>next
 slow>next = slow>next>next
 else
 slow>next = nil
 end
return head
C++ solution
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* fast;
ListNode* slow;
fast = head;
slow = head;
while(n){
fast = fast>next;
n;
}
if(fast == NULL){
head = head>next;
return head;
}
while(fast>next){
slow = slow>next;
fast = fast>next;
}
if(slow>next && slow>next>next){
slow>next = slow>next>next;
} else {
slow>next = NULL;
}
return head;
}
};
Golang solution
func removeNthFromEnd(head *ListNode, n int) *ListNode {
node := &ListNode{}
node.Next = head
slow, fast := node, node
for ; n > 0; n {
fast = fast.Next
}
for ; fast.Next != nil; slow, fast = slow.Next, fast.Next {}
slow.Next = slow.Next.Next
return node.Next
}
Javascript solution
var removeNthFromEnd = function(head, n) {
let fast = head;
let slow = head;
while(n > 0) {
fast = fast.next;
n;
}
if(fast === null) return head.next;
while(fast.next !== null) {
slow = slow.next;
fast = fast.next;
}
slow.next = slow.next.next;
return head;
};
Dry Run
Let's dryrun our algorithm.
head = [1, 2, 3, 4, 5]
n = 2
Step 1: fast = head, slow = head
slow, fast  [1, 2, 3, 4, 5]
Step 2: Loop while n > 0
2 > 0 = true
fast = fast>next
fast

slow  [1, 2, 3, 4, 5]
n
n = 1
Step 3: Loop while n > 0
1 > 0 = true
fast = fast>next
fast

slow  [1, 2, 3, 4, 5]
n
n = 0
Step 4: Loop while n > 0
0 > 0 = false
Step 5: if fast == nil
= false
Step 6: Loop while fast.next != nil
= true
// fast.next pointing to node 4 address
slow = slow.next
fast = fast.next
slow fast
 
[1, 2, 3, 4, 5]
Step 7: Loop while fast.next != nil
= true
// fast.next pointing to node 5 address
slow = slow.next
fast = fast.next
slow fast
 
[1, 2, 3, 4, 5]
Step 8: while fast.next != nil
= false
Step 9: if slow.next && slow.next.next
slow is node 3
slow.next is node 4
slow.next is node 5
slow.next = slow.next.next
// so node 3 next is now pointing to 5
Step 10: return head
[1, 2, 3, 5]