CareerCup

you just leaked the memory of the 5th node

The above is the correct solution, there is no need to move from your current pointer.

The only problem is the memory management. If you are in C++, then you just left that memory block allocated and have no way to reference it for the delete function to clean it up. If you are using Java, then the de-referencing will mark the 4th node for garbage collection.

This would be a very excellent call out to demonstrate your understanding of appropriate memory management in your given language.

@masterjaso: @uruk_hai is correct.He just leaked the memory of the 5 node basically making it a dangling pointer.

Please explain the use of XOR (^) operator.

could you please explain the Xor logic.
thanks
sanjay

Once you do a free, ptr->next->next->next can be junk. So wrong solution.

Yes, convoluted thinking.

Equivalent code without the convoluted XOR is

free(ptr->next->next);
ptr->next->next = ptr->next->next->next;

This assumes that the linked list nodes are contiguous and it also does not repair the linked list when the node is deleted.

whoops just realized i scrogged line 2 with line 1's assignment

second line should be removed