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Dracula: Dead and Loving It
1995 · Film · 88 min
Directed by Mel Brooks
When a lawyer shows up at the vampire's doorstep, he falls prey to his charms and joins him in his search for fresh blood. Enter Professor Van Helsing, who may be the only one able to vanquish the Count.
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> They made a mistake and got too greedy and make a subscription I barely paid attention to something that got too expensive...
That's the thing, in the past I I'd go months without watching anything on Netflix yet still pay them for it, so they were essentially getting a free $8 a month (or whatever) off me during that time.
Now the prices have gone up so much it doesn't make sense to keep subscribed for something I'm barely using, just so I can easily watch a single show that interests me once every three months or so.
The only streaming service that has makes some sense for me to pay for right now is Youtube's, since that's my goto for videos anyway, and it's nice not to have ads for those.
Additionally they also have a bunch of great older movies free to watch movies on Youtube, probably a better selection (of movies I care about) than Netflix right now (saw/rewatched Train to Busan, Little Shop of Horrors, Ghostbusters, It's a Mad, Mad, Mad, Mad World, Tropic Thunder, Fifth Element, Big Trouble in Little China, Dr. Strangelove, 12 Angry Men, Pee-wee's Big Adventure, The Truman Show, Air Force One, Dirty Work, and Dracula: Dead and Loving It recently).
You're on the right track with the recursive dictionary search.
For the Sling Blade Runner problem, I would make a trie (http://en.wikipedia.org/wiki/Trie) out of hash tables, like this:
trie = { 'dracula': {'': None, 'dead': {'and': {'loving': {'it': "Dracula: Dead and Loving It!"}}}}, 'the': {'brides': {'of': {'dracula': "The Brides of Dracula"}}}, # ... }
Then, for each word in each movie title, walk the trie from the beginning until you run out of matches (no match), or run out of words (at least one match).
Your anagram server is similar, but since they're not ordered like movie titles, you have to walk your scrambled word(s) and only generate permutations that have a matching trie path.
It's also quite possible to do it for phrases, not just words.
You can cheat a little by representing your anagram as a 26-key multiset containing the number of times each letter appears (in Python, use a dict; in C, use an array of 26 integers).
For each letter that appears in your anagram multiset one or more times, make a copy of the original multiset, and recursively descend the Dictionary trie, removing each matching letter from a new copy of the multiset (they're tiny, so don't worry). For each complete word you find, add it to a list of result candidates, along with the multiset representing the remaining letters in the anagrammatic phrase.
Repeat the process for each result candidate, growing your list into a tree. Stop when you run out of matches, or when you run out of letters in your multisets. (You can set a maximum number of leftover letters to accept, if you're having trouble matching a phrase).
If you test the letters in alphabetical order, your result phrases will already be sorted.