19. Remove Nth Node From End of List

//🎯DAY 13 PROBLEM 2
//The approach is very simple i.e. we use two pointers, move the first pointer till n and then move 2 pointers simultaneously.. When the second pointer moves till end, the first pointer
//will reach nth position from the end. We just need to handle some edge cases:

/**
 * 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) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(head->next==NULL)
            head=NULL;
       else
       {
      ListNode *p1=head,*p2=head,*prevp1=head;
        int count=0;
        while(count<n-1)
        {
            p2=p2->next
            count++;
        }
        //📌Take care of this when we have to remove the head of the linked list itself then the process would be different and if we follow the same process,
        //then we null->next is bound to occur and error occurs
           if(p2->next==NULL)
           {
               head=head->next;
               return head;
           }
      while(p2->next!=NULL)
      {
          prevp1=p1;
          p1=p1->next;
          p2=p2->next;
      }
        prevp1->next=p1->next;
     
      }
           return head;
    }
};