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@gcapell / forked from okaq/

Solution: Runs (Google Code Jam 2011 World Finals Problem A)
"" Solution: Runs (Google Code Jam 2011 World Finals Problem A)
"" by:
"" run: python infile outfile
import sys
import time
import collections
MOD = 1000003
def bc(n, k):
"""Binomial coefficient"""
r = 1
for i in range(1, k+1):
r = r * (n-i+1) / i
return r
bcmd = {}
def bcm(*args):
"""binomial coefficient (memoized)"""
return bcmd[args]
except KeyError:
bc0 = bc(*args)
bcmd[args] = bc0
return bc0
def count_transitions(existing_letters, existing_runs, add_letters, final_runs):
Given existing_letters (count) arranged in existing_runs (count),
in how many ways can we add add_letters to form final_runs?
if existing_runs == 0:
return 1 if final_runs ==1 else 0
# We can add letters at boundaries (adds one run)
# or within existing runs (adds two runs).
return sum (
# How many ways can we insert at a boundary?
bcm(existing_runs + 1, at_boundary) *
# How many ways can we insert within an existing run?
bcm(existing_letters - existing_runs, in_run) *
# How many ways can we arrange the letters we are adding
# within our (at_boundary+in_run) insertion points?
bcm(add_letters - 1, at_boundary + in_run - 1)
) for in_run, at_boundary in gen_insertions(final_runs - existing_runs)
def gen_insertions(delta):
for in_run in range(delta/2 + 1):
yield in_run, delta - 2 * in_run
def solve(freq, runs_goal):
Dynamic programming algo that runs iteratively up to original number of runs.
Complicated, but fast
runs_count = [1] + [0] * (runs_goal-1)
existing_letters = 0
for add_letters in freq:
runs_count_prev = list(runs_count)
runs_count = [0]
for r in range(1, runs_goal+1):
count_transitions(existing_letters, r0, add_letters, r) * runs_count_prev[r0]
for r0 in range(r)
) % MOD)
existing_letters += add_letters
return runs_count[-1]
def get_words(filename):
"""Yield (testnum, word) pairs read from input filename"""
with open(filename) as fp:
ntests = int(fp.readline())
for testnum in range(1, ntests+1):
yield testnum, fp.readline().strip()
def analyse(word):
Given word return (freq, runs) pair
where freq is list of counts of letters,
runs is number of runs in word.
prev = None
runs = 0
hist = collections.defaultdict(int)
for c in word:
hist[c] += 1
if c != prev:
runs += 1
prev = c
return hist.values(), runs
class Timer:
"""Simple split timer"""
def __init__(self):
self.start = time.time()
def split_time(self):
return "%.3f" % (time.time() - self.start)
def main():
t = Timer()
with open(sys.argv[2], 'w') as fout:
for testnum, word in get_words(sys.argv[1]):
freq, runs = analyse(word)
s0 = solve(freq, runs)
fout.write("Case #%d: %d\n" % (testnum, s0))
print "split time for case #%d (%d): %s" % (testnum, s0, t.split_time())
print "total time: %s" % (t.split_time())
if __name__ == '__main__':
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