# Efficiency on Project Euler Problem #8

I am doing Project Euler problem #8:

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?

Here is my code to this problem:

L = []
L.append("73167176531330624919225119674426574742355349194934")
L.append("96983520312774506326239578318016984801869478851843")
L.append("85861560789112949495459501737958331952853208805511")
L.append("12540698747158523863050715693290963295227443043557")
L.append("66896648950445244523161731856403098711121722383113")
L.append("62229893423380308135336276614282806444486645238749")
L.append("30358907296290491560440772390713810515859307960866")
L.append("70172427121883998797908792274921901699720888093776")
L.append("65727333001053367881220235421809751254540594752243")
L.append("52584907711670556013604839586446706324415722155397")
L.append("53697817977846174064955149290862569321978468622482")
L.append("83972241375657056057490261407972968652414535100474")
L.append("82166370484403199890008895243450658541227588666881")
L.append("16427171479924442928230863465674813919123162824586")
L.append("17866458359124566529476545682848912883142607690042")
L.append("24219022671055626321111109370544217506941658960408")
L.append("07198403850962455444362981230987879927244284909188")
L.append("84580156166097919133875499200524063689912560717606")
L.append("05886116467109405077541002256983155200055935729725")
L.append("71636269561882670428252483600823257530420752963450")

M = []

for s in L:
for i in list(s): M.append(i)

greatest_product = 0

for i in range(len(M)):
product = 1
substring = M[i:i+13]
for digit in substring:
product*=int(digit)
if product > greatest_product:
greatest_product = product

print greatest_product


The code works, but how would I improve the efficiency?

• The loop iterates over the whole range(len(M)). It means that the last few products exercise less than 13 digits. It doesn't matter in this particular example (they are all zeros anyway), but may lead to erroneous result in general case. Iterate over range(len(M) - 13).

• Digit-to-int conversion happens 13 times for each digit. Better convert them once in a separate loop before doing calculations.

• Every time a 0 is encountered, you may safely skip the whole bunch of digits: the product will be 0 anyway.

• A sliding window technique may help.

In pseudocode:

product *= data[head]
product /= data[tail]
tail += 1


(be careful to not divide by 0).

I would get rid of the nested for loop as it is not needed.

If you're familiar with the problem of finding $k$th last element in the single linked list then you can borrow same idea for this problem.

Hint: let's say you are checking some 13 digits starting at index $i$ till $i+12$, and its product is $P$. To check the product of the next combination of 13 digits, you just need to divide $P$ by $i$th digit and multiply it by $i+13$th digit, and so on, just keep in mind: edge cases.

• Welcome to Code Review! You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and how it improves upon the original) so that the author can learn from your thought process. – Quill Aug 13 '15 at 8:03

As the question was about improving the efficiency I'd like to point out that the search can be limited to smaller strings.

As was pointed out the 0s will always produce a product of 0. Keeping that in mind you can see that you only need to search the string between the zeros.

Following this train of thought further optimizations come naturally:

• only strings longer than 13 digits can apply
• as these strings no longer contain zeros the sliding window suggested by @vnp will work without checking for 0