Faster than 100% C++ code ✔
idea is to take two pointers and traverse the first pointer n nodes\ntill the time 2nd pointer will be as it is\nnow when you\'ll traverse the both poniters sim
idea is to take two pointers and traverse the first pointer n nodes\ntill the time 2nd pointer will be as it is\nnow when you\'ll traverse the both poniters simultaneously then definitely first pointer\nis gonna be at the position last and 2nd pointer will be just before the node we\'ve to delete\n\nhere first pointer is fast and 2nd pointer is slow; \nhere is the code implementation os the above approach------------------------\n```\nListNode* removeNthFromEnd(ListNode* head, int n) {\n        ListNode *start=new ListNode(); \n        start->next=head;\n        ListNode *slow=start,*fast=start;\n        for(int i=0;i<n;i++)\n            fast=fast->next; //traverse n nodes\n        \n        while(fast->next!=nullptr){  //as i told fast will be at the last node\n            fast=fast->next;\n            slow=slow->next;\n        }\n        ListNode *del=slow->next; //step to finish matlab khatam krna hai node ko\n        slow->next=slow->next->next; //next pointer of the slow will point on just after the victim node\n        delete(del); //Khatam \uD83D\uDC7B\n        return start->next;\n    }\n```\n\nmy english ;P\n\uD83D\uDE42 why you guys always forget to upvote, it motivates me to post such answers.\n<a href="https://cutt.ly/KalyanChannel">You can checkout the Youtube channel \uD83D\uDCF1  </a>
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