Tag Archives: one-liner

project euler problem #8

Aside

Find the greatest product of five consecutive digits in the following 1000-digit number:

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

http://projecteuler.net/problem=8

highlight below for my solution:


a_string = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"

print max([reduce(lambda x, y : x * y, [int(x) for x in a_string[i:i+5]]) 
           for i in range(len(a_string) - 5)
          ])

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project euler problem #1

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If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.

http://projecteuler.net/problem=1

  • try solving it as a python one-liner
  • try optimizing your solution for speed

highlight below for my solution:


sum(range(0,1000,3)) + sum(range(0,1000,5)) - sum(range(0,1000,15))

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python counters

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in my pythonic trolling of the internet, i found a character counter, written by ashoksk. given a string of text, return a count for each character in the string (a tool that might be used when attempting to crack an encrypted message)

so i wondered, could this be a python one liner? turns out that, from python 2.7 on, it can be (highlight to see the code below):


from collections import Counter
def fn(x): c = Counter(x); return {x: c[x] for x in c.keys()}

it’s a cheat, though. anyone have a real one-line implementation of this?

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recursive touch

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i just moved a bunch of media files from one device, to the other, and the file creation dates for a small number of my directories (and their contents) moved a few years into the future.

this problem smelled like a python 1 liner; i came pretty close – highlight to see the code:


import os, sys, subprocess

for d,s,fs in os.walk(sys.argv[1]):
    for p in (s+fs):
        subprocess.call(["touch",os.path.join(d,p)])

(source online, in case I improve it)

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riddle 2 (python)

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Count the number of days between two days.

same as last time, highlight to see answer(s), in python, below. 


def fn(*args): 
      x = '/'.join(str(i) for i in args[:3])
      y = '/'.join(str(i) for i in args[3:])
      return abs((datetime.strptime(x,"%Y/%m/%d")-datetime.strptime(y,"%Y/%m/%d")).days)

fn(2001,1,1,2002,2,2)

#depending on input format, could do a one-liner:

def fn(x,y): abs((datetime.strptime(x,"%Y/%m/%d")-datetime.strptime(y,"%Y/%m/%d")).days)

#it's probably possible to replace "%Y/%m/%d" with "%c"

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