Given a singly-linked list, devise a time and space efficient algorithm to find the mth-to-last element of the list. Implement your algorithm, taking care to handle relevant error conditions. Define mth to last such that when m = 0, the last element of the list is returned.

mfromlast.cpp

Element *findMToLastElement( Element *head, int m ) {
	if (!head || n < 0) {
		return NULL;
	}
	Element *first, *second = head, head;
	for(int i = 0; i < m; i++) {
		first = first->next;
		if (!first) {
			return NULL;
		}
	}

	while(first->next) {
		first = first->next;
		second = second->next;
	}
	return second;
}