The following code is a modification from section $5\mathrm{.}3$ which is moving the list$\_$head$\_$remove function call two lines ealier. What will go wrong?

bag::size$\_$type bag::erase$($const value$\_$type& target$)$

$//$ Library facilities used: cstdlib, node$1.$h

$\{$

size$\_$type answer $=0$;

node $*$target$\_$ptr;

target$\_$ptr $=$ list$\_$search$($head$\_$ptr$,$ target$)$;

while $($target$\_$ptr $!=$ NULL$)$

$\{$

$//$ Each time that target$\_$ptr is not NULL, we have another occurrence of target.

$//$ We remove this target using the same technique that was used in erase$\_$one.

target$\_$ptr$->$set$\_$data$($ head$\_$ptr$->$data$())$;

list$\_$head$\_$remove$($head$\_$ptr$)$;

target$\_$ptr $=$ target$\_$ptr$->$link$()$;

target$\_$ptr $=$ list$\_$search$($target$\_$ptr$,$ target$)$;

$--$many$\_$nodes;

$++$answer;

$\}$

return answer;

$\}$

Group of answer choices

The problem occurs when the target is the second item on the linked list. In this case, the target pointer is at the head of the list, so it would be a mistake to remove the head node before moving the target pointer forward

The problem occurs when the target is the first item on the linked list. In this case, the target pointer is at the head of the list, so it would be a mistake to remove the tail node before moving the tail pointer backward.

The problem occurs when the target is the first item on the linked list. In this case, the target pointer is at the head of the list, so it would be a mistake to remove the tail node before moving the tail pointer backward.

The problem occurs when the target is the last item on the linked list. In this case, the target pointer is at the tail of the list, so it would be a mistake to remove the tail node before moving the tail pointer backward.

The problem occurs when the target is the first item on the linked list. In this case, the target pointer is at the head of the list, so it would be a mistake to remove the head node before moving the target pointer forward.