Project Euler: Problem 8

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

(see the number below)

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

Use NumPy for this computation



In [2]:

    
import numpy as np



In [3]:

    
d1000 = """
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
"""

First, I replaced all the \n's in d1000 with a space to have just a string of digits



In [4]:

    
d1000_new = d1000.replace('\n', '')

Then, I appended each integer value in d1000_new to a new list



In [5]:

    
lst = []
for i in d1000_new:
    lst.append(int(i))

Then I made this list into an array



In [6]:

    
a = np.array(lst)

I started the main code by defining the maximum product as zero, and then a for loop that looped through all the digits in the list of digits. Using an if statement, I set a bound so no groups of thirteen would attempted to be made where no values would be found(like near the end of the number). I used np.prod to calculate the product of all thirteen digit groups using array slicing. A final if statement goes through the products as they're created, and if they're larger than the maximum the previous maximum product, then that will be the new maximum product. Finally, I print the max product.



In [12]:

    
maximum_prod = 0
for n in range(0, len(a)+1):
    if n <= (len(a) - 12):
        prod = np.prod(a[n:n+13])
        if prod > maximum_prod:
            maximum_prod = prod
print(maximum)

23514624000



In [26]:

    
assert True # leave this for grading