matching end of string

Question

I'm looking for the best most efficient way to match the end of a single string with a value from a predefined list of strings.
Something like

my_str='QWERTY'
my_lst=['QWE','QQQQ','TYE','YTR','TY']  

match='TY' or match=['TY']

Under the restrictions

len(my_lst) is known but arbitrary thus could be very long, probably around 30
elements in my_lst may have different len so I can't just check a defined last portion of my_str every time
for my_str as well as the matching elements in my_lst they can be either strings or lists, whichever is more efficient (see background)
len(my_str) is mostly small, no longer than 8 characters
in function wouldn't do as I need the matching to occur exclusively at the end.
endswith is no use on it's own since it would only return a Boolean
the match should always be unique or [] as no elements in my_lst would share ending with one another

little background may skip
I started with this problem as a list problem such as ['Q','W','E','R','T','Y'] where I would have a list of lists of 1 character strings for the matching and I was thinking of running a reverse iteration as [::-1] for the checking for every candidate.
Then I realized it was possible to concatenate the inner lists since they contained only strings and run the same logic on the resulting strings.
Finally I came across the endswith string method reading this question but it wasn't quite what I needed. Furthermore my problem can't be generalized to be solved with os module or similar since it's a string problem, not a pathing one.
end of background
I made my approach in this two ways

match=filter(lambda x: my_str.endswith(x), my_lst)
match=[x for x in my_lst if my_str.endswith(x)]

I succeeded but I would like to know if there is some built-in or best way to find and return the matched ending value.

Thanks.

Answer 1

Here's a way using a trie, or prefix tree (technically a suffix tree in this situation). If we had three potential suffixes CA, CB, and BA, our suffix tree would look like

     e
    / \
  A     B
 / \    |
B   C   C

(e is the empty string) We start at the end of the input string and consume characters. If we run across the beginning of the string or a character that is not a child of the current node, then we reject the string. If we reach a leaf of the tree, then we accept the string. This lets us scale better to very many potential suffixes.

def build_trie(suffixes):
    head = {}
    for suffix in suffixes:
        curr = head
        for c in reversed(suffix):
            if c not in curr:
                curr[c] = {}
            curr = curr[c]
    return head

def is_suffix(trie, s):
    if not trie:
        return True
    for c in reversed(s):
        try:
            trie = trie[c]
        except KeyError:
            return False
        if not trie:
            return True
    return False

trie = build_trie(['QWE','QQQQ','TYE','YTR','TY'])

gives us a trie of

{'E': {'W': {'Q': {}}, 
       'Y': {'T': {}}},
 'Q': {'Q': {'Q': {'Q': {}}}},
 'R': {'T': {'Y': {}}},
 'Y': {'T': {}}}

If you want to return the matching suffix, that's just a matter of tracking the characters we see as we descendt he trie.

def has_suffix(trie, s):
    if not trie:
        return ''
    letters = []
    for c in reversed(s):
        try:
            trie = trie[c]
            letters.append(c)
        except KeyError:
            return None
        if not trie:
            return ''.join(letters)
    return None

It's worth noting that the empty trie can be reached by both build_trie(['']) and build_trie([]), and matches the empty string at the end of all strings. To avoid this, you could check the length of suffixes and return some non-dict value, which you would check against in has_suffix