Problem

Given the head of a linked list, remove the n<sup>th</sup> node from the end of the list and return its head.

List: [1] -> [2] -> [3] -> [4] -> [5]

Value of N: 2

Output: [1] -> [2] -> [3] -> [5]

Given Code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:

Solution

Logical

Let's break down this code in simpler terms:

1. Check if n is 0:

    if n <= 0:
        return head
   

   This part checks if the value of n is zero. If it is, the function immediately returns the original list (head) without making any changes.

2. Calculate the Length of the List:

    curr = head
    len = 0
    while curr:
        len += 1
        curr = curr.next
   

   This section counts how many nodes are in the given linked list (head). It does this by moving through the list node by node and keeping track of the count (len).

3. Check if n is Greater than the Length of the List:

    if n > len:
        return None
   

   If the value of n is greater than the length of the list, the function returns None because it means we can't remove the n-th node from the end of the list if the list is shorter than n.

4. Determine the Position of the Node to Remove from the Beginning:

    nodeFromFirst = len - n + 1
   

   It calculates the position of the node that needs to be removed from the beginning of the list. For example, if n is 2, and the list has 5 nodes, it calculates the position of the node to remove as 5 - 2 + 1 = 4.

5. Initialize Two Lists:

    returnList = updateList = ListNode()
   

   It initializes two lists (returnList and updateList) with an empty node. These lists will be used to construct the modified list.

6. Remove the N-th Node from the End:

    for i in range(nodeFromFirst):
        if i == nodeFromFirst - 1:
            returnList.next = head.next
            break
        else:
            returnList.next = ListNode(head.val)
            returnList = returnList.next
            head = head.next
   

   This loop iterates through the original list up to the position calculated earlier (nodeFromFirst). It creates a new list (returnList) excluding the node to be removed. The loop uses a temporary list (updateList) to keep track of the modified list.

7. Return the Modified List:

    return updateList.next
   

   Finally, it returns the modified list starting from the node after the initially created empty node (updateList).

In simpler terms, this code removes the N-th node from the end of a linked list. It first checks if n is zero or greater than the length of the list. Then, it calculates the position of the node to remove, creates a new list without that node, and returns the modified list.

Code

class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        if not bool(n):
            return head
        curr = head
        len = 0
        while curr:
            len += 1
            curr = curr.next
        if n > len:
            return None
        nodeFromFirst = len - n + 1
        returnList = updateList = ListNode()
        for i in range(nodeFromFirst):
            if i == nodeFromFirst - 1:
                returnList.next = head.next
                break
            else:
                returnList.next = ListNode(head.val)
                returnList = returnList.next
                head = head.next
        return updateList.next