find_m_to_last_llist.py

# Given a linked list, this method
# will return m'th element to the last
# 2->3->4->8->5; m=2 will return 8
# since 8 is second to last

from linked_list_data_structure import LinkedList


def findMToLast(l_list, m):
    current = l_list.head
    count = 0

    while current is not None and count < m:
        count += 1
        current = current.getNextNode()

    m_behind = l_list.head
    while current.next_node is not None:
        current = current.getNextNode()
        m_behind = m_behind.getNextNode()

    return m_behind


linked_list = LinkedList()
m_to_last = 3
# Returns the third element from last
print(findMToLast(linked_list, m_to_last))