embedly challenge

embedly.py

# Problem 1
fact = lambda n: reduce(lambda x,y: x*y, range(1, n+1), 1)
convert = lambda n: sum(map(int,str(n)))

def p1gen():
    i = 0
    while True:
        yield (i,convert(fact(i)))
        i+=1

p = p1gen()
result = (0,0)
while result[1] != 8001:
    result = p.next()

print("Problem 1: "+str(result[0])) # 787

# or

l = [ x for x in range(0,1000) if convert(fact(x))==8001 ]
print("Problem 1: "+str(min(l))) # 787

# Problem 2
from HTMLParser import HTMLParser
import numpy as np

class P2Parser(HTMLParser):
    def __init__(self):
        HTMLParser.__init__(self)
        self.pdepths = []
        self.depth = -3         # account for head/body
    def handle_starttag(self, tag, attrs):
        # print(str(tag)+" "+str(self.depth))
        self.depth += 1
        if(tag=="p"):
            self.pdepths.append(self.depth)
    def handle_endtag(self, tag):
        self.depth -= 1
parser = P2Parser()
parser.feed(open("2.html", 'r').read())
print("Problem 2: "+str(np.std(parser.pdepths))) # 1.411...

# Problem 3

from itertools import *
s=2520.0
l = [s/x for x in range(1,901)]
half_of_words = sum(l)/2.0
# ugly, but I only had < 1 hr to work on these problems...
print("Problem 3: "+str(len(list(takewhile(lambda x: x < half_of_words,
                                           np.cumsum(l))))+1)) # 22

Problem 3

I think your implementation is not correct.

>>> s=2520.0
>>> l = [s/x for x in range(1,900)]
>>> len(l)
899

length of l must be 900, as the problem states "Given that the text has 900 unique words".

Very possibly--there was some ambiguity as to whether the question wanted to overshoot or undershoot (by 1) the halfway point so I just picked the one that went over (and didn't thoroughly examine the consequences--since it only makes a difference of ~3 words, it's not surprising it still gives the same answer).

I wonder what is that +1 you are adding at the end?

print("Problem 3: "+str(len(list(takewhile(lambda x: x < half_of_words, np.cumsum(l))))+1)) # 22

I made the assumption that the problem wanted to overshoot the halfway point (i.e. the first element > half_of_words) rather than undershoot (and when I checked the answer against the webapp, this is indeed what was wanted). But, thanks to the lambda I passed to takewhile, it will return all the elements except the last element that puts the number of elements over the halfway point. So I added this final element. If itertools had a "takeuntil" function, I would have used that instead, but as I was coding for speed, I just used what was in the libraries.