# UNIT 2 Problem Set - Question 6
# Define a procedure, find_last, that takes as input
# two strings, a search string and a target string,
# and returns the last position in the search string
# where the target string appears, or -1 if there
# are no occurrences.
#
# Example: find_last('aaaa', 'a') returns 3

# Make sure your procedure has a return statement.

def find_last(search, target):
    search_len = int(len(search))
    target_len = int(len(target))
    if (search.find(target)) == (-1):
        return -1
    search = search[::-1]
    target = target[::-1]
    last_char = search.find(target)
    flip_first = last_char + target_len
    first_char = search_len - flip_first
    return first_char

sentence = "when I go walking, after midnight, out in the starlight"
target = "ight"
print (find_last(sentence,target))

#note str length is no. characters whereas position no. character starts at zero
#note above approach reverses both strings
#first char last occurrence target str, if reversed, is last char first occurrence in reversed search str
#easy to find first character first occurence of reversed target str in reversed search string
#then, arithmetic using string lengths
#add target str length to position no. first char reversed target str first occurrence in reversed search str
#logic is position no. in reversed str deducted from search str length gives position no. in original search str