LinkedList Questions: [Two Pass] Delete nth node from end

In this series of posts, I will discuss coding questions on the LinkedList Data structure.
The posts in this series will be organized in the following way,

Question Link ❓
Possible Explanation 📝
Documented C++ Code 🧹
Time and Space Complexity Analysis ⌛🌌

The Question

Given the head of a linked list, remove the nth node from the end of the list and return its head.

Constraints:

The number of nodes in the list is sz.
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

💡 Give yourself atleast 15-20 mins to figure out the solution :)

There are two approaches possible, in this post we will see the first one.

Approach 1: Two Pass

If you think a bit, nth node from end is list_len - n + 1 th node from beginning.

So our algorithm is:

Find the length of LinkedList → L
Delete the (L - n + 1)th node from beginning.

C++ Code

Definition of LinkedList

//Definition for singly-linked list.
struct ListNode
{
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
};

Solution

ListNode *removeNthFromEnd(ListNode *head, int n)
    {
        //- if LL is empty
        if (!head)
            return head;

        //note: first pass : O(n)
        //- getting length of LL 
        int cnt = 0;
        ListNode *temp = head;
        while (temp)
        {
            cnt++;
            temp = temp->next;
        }

        //* Standard Procedure to delete (k+1)th node from beginning

        //note: Required Node: (cnt - n +1)th node
        //note: we have to go "cnt-n" times deep to stand at required node
        int k = cnt - n;
        ListNode *cur = head;
        ListNode *prev = nullptr; //it will point one node preceding to cur

        //note: second pass :O(n)
        while (k > 0)
        {
            prev = cur;
            cur = cur->next;
            k--;
        }

        //- first node of the LL is to be deleted
        if (!prev)
        {
            temp = cur;
            cur = cur->next;
            delete temp;
            head = cur; //! cur is the new head
        }
        else
        {
            prev->next = cur->next;
            delete cur;
        }
        return head;
    }