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Programming Challange from Erann Gat: http://www.flownet.com/ron/papers/lisp-java/
# Josiah Carlson - Programming Challange from Erann Gat:
# http://www.flownet.com/ron/papers/lisp-java/
# Given a list of words and a list of phone numbers, find all the ways that
# each phone number can be expressed as a list of words.
from collections import defaultdict
import sys
MAPPING = {'e':0, 'j':1, 'n':1, 'q':1, 'r':2, 'w':2, 'x':2, 'd':3, 's':3,
'y':3, 'f':4, 't':4, 'a':5, 'm':5, 'c':6, 'i':6, 'v':6, 'b':7, 'k':7,
'u':7, 'l':8, 'o':8, 'p':8, 'g':9, 'h':9, 'z':9}
DIGITS = set(map(str, range(10)))
def build_lookup(wordlist):
# We're going to build a structure as follows:
# {3 : {783: ['bo"s'], 107: ['neu'], ...}, ...}
# This structure allows for the fast lookup of a particular phone number
# of a given length.
# We could have used {(3, 783): ...} instead, but creates an extra tuple
# that is unnecessary.
lookup = defaultdict(lambda:defaultdict(list))
for word in wordlist:
# Filter those digits that we care about.
encoded = [str(MAPPING[ch]) for ch in word.lower() if ch in MAPPING]
if not encoded:
continue
# We're going to exploit Python's fast int() function instead of
# manually constructing the int ourselves. We do this for the sake of
# saving space, though it's probably unnecessary.
lookup[len(encoded)][int(''.join(encoded), 10)].append(word.strip())
return lookup
def get_numbers(number):
# We're going to generate the lookups for possible matches from longest to
# shortest
x = [(0, 0)]
for v in number:
x.append((x[-1][0] + 1, x[-1][1] * 10 + v))
x.reverse()
x.pop()
return x
def find_matches(number, lookup, already_skipped=False):
# We're going to generate possible match lookup keys from longest to
# shortest, then if we find a match, recurse. If we don't find any
# matches, we'll recurse with a digit as a filler if we've not already
# done so at the most recent level.
# When the recursive calls yield something of value, we'll append it to
# what we've found, and yield what we've found.
if not number:
# Return the base case of an empty match, which will allow for
# callers to add prefixes.
yield ()
return
found = False
for used_digits, value in get_numbers(number):
# Find the matching words, if any.
words = lookup.get(used_digits, {}).get(value, ())
sk = not words
found = found or bool(words)
# If we can skip a digit, we'll try to do so now, and filter later.
if sk and used_digits == 1 and not already_skipped and not found:
# Use a digit as a filler.
words = (str(value),)
elif not words:
# No words or not possible for a filler, skip this entry
continue
# As long as we can skip, or we've found a match here, we'll recurse.
for match in find_matches(number[used_digits:], lookup, sk):
for word in words:
yield (word,) + match
def build_and_find_all(dictionary_file, number_file):
lookup = build_lookup(open(dictionary_file))
for line in open(number_file):
line = line.strip()
# Filter our digits...
number = [n for n in line if n in DIGITS]
if not number:
continue
# Find our matches...
for match in find_matches(map(int, number), lookup):
if match:
# Properly print our matches.
print "%s: %s"%(line, ' '.join(match))
if __name__ == '__main__':
build_and_find_all(*sys.argv[1:])
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