Given an singly linked list and poisition "n", we have to remove the nth item from end
Answer: In either of implementation, the complexity will be O(n).
Steps of the first questions are as:
1) Take a temporary pointer.
2) When you cross the (n)th element, store the address of the nth position in the temporary pointer or reference such that traversal pointer must be having (n+1)th address.3) Now after above step, always update the value of the temp by the new address or next address in the singly linked list.
4) After reaching at the end, temp pointer will be in the (n+1)th position from the end.
5) Now just replace the value from the (n-1)(from the end) address to get the new linked list.
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I hope I answered your question well in a very economical way.
Second answer will be O(nlogn) because of one standard sort.
2. Second step will be mere replacing the duplicate entries and that will be of order n hardly.