Project Euler Problem 8

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

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?

There's not too much to this problem. We can slide a window of length num_digits across the string, compute the product of digits in each window, and track the maximum. One optimization is to skip any window containing a '0' since its product will be zero.

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

function product_of_digits(str)
    prod = 1
    for c in str
        digit = parse(Int, c)
        prod *= digit
    end
    return prod
end

function largest_product_in_series(num_digits)
    max_product = 0
    max_substring = ""

    for i in 1:(length(BIG_NUMBER) - num_digits + 1)
        substring = BIG_NUMBER[i:(i + num_digits - 1)]

        if '0' in substring
            continue
        end

        product = product_of_digits(substring)

        if product > max_product
            max_product = product
            max_substring = substring
        end
    end

    return max_product, max_substring
end

Calling largest_product_in_series(13) returns the solution in 20.751 μs. We can also run the example from the problem description, largest_product_in_series(4), which returns $(5832, \text{"9989"})$ in 19.800 μs.

The algorithm runs in $\mathcal{O}(nk)$ time where $n$ is the length of the string and $k$ is the window size. The early termination on zeros provides a constant-factor speedup but doesn't change the asymptotic complexity.