The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

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

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?



In [1]:

    
from six.moves import map, range, reduce

Version 1 - Sliding Window

Let's first set a variable to the number



In [2]:

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



In [3]:

    
n









    Out[3]:





7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450

Now let's form an array with elements representing the digits of the number.



In [4]:

    
num_to_list = lambda n: list(map(int, list(str(n))))

This solution is kind of hacky and relies on specific characteristics of these Python built-in functions. Ideally, we'd build an array from the integer itself without all this type casting. But this will do for now.



In [5]:

    
num_to_list(n)[:10]









    Out[5]:





[7, 3, 1, 6, 7, 1, 7, 6, 5, 3]

Now we need to iterate over this array with some kind of sliding "window" of length $L=13$.



In [6]:

    
window = lambda lst, n: map(list, zip(*[lst[i:-n+i] for i in range(n)]))

Example: Partial function application and creating a list of odd numbers



In [7]:

    
pairwise = lambda lst: window(lst, 2)
[x+y for x, y in pairwise(range(10))]









    Out[7]:





[1, 3, 5, 7, 9, 11, 13, 15]



In [8]:

    
list(window(num_to_list(n), 13))[:10]









    Out[8]:





[[7, 3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3],
 [3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 0],
 [1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 0, 6],
 [6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 0, 6, 2],
 [7, 1, 7, 6, 5, 3, 1, 3, 3, 0, 6, 2, 4],
 [1, 7, 6, 5, 3, 1, 3, 3, 0, 6, 2, 4, 9],
 [7, 6, 5, 3, 1, 3, 3, 0, 6, 2, 4, 9, 1],
 [6, 5, 3, 1, 3, 3, 0, 6, 2, 4, 9, 1, 9],
 [5, 3, 1, 3, 3, 0, 6, 2, 4, 9, 1, 9, 2],
 [3, 1, 3, 3, 0, 6, 2, 4, 9, 1, 9, 2, 2]]



In [9]:

    
prod = lambda iterable: reduce(lambda x, y: x*y, iterable)

Example: Compute $8!$



In [10]:

    
prod(range(1, 9))









    Out[10]:





40320



In [11]:

    
list(map(prod, window(num_to_list(n), 13)))[:20]



In [12]:

    
max(map(prod, window(num_to_list(n), 13)))









    Out[12]:





23514624000



In [13]:

    
max(map(prod, window(num_to_list(n), 4)))









    Out[13]:





5832

Given the frequency of 0's when applying the sliding window, the next optimization to consider is skipping over windows containing at least one zero.



In [14]:

    
from collections import Counter
Counter(map(prod, window(num_to_list(n), 13))).most_common(10)









    Out[14]:





[(0, 724),
 (6718464, 3),
 (313528320, 3),
 (28449792, 3),
 (9797760, 3),
 (371589120, 3),
 (495452160, 3),
 (928972800, 3),
 (235146240, 2),
 (78382080, 2)]