Euler 0008

The Problem:

In this giant 1000 digit number the greatest product of 4 adjacent numbers is 9*9*8*9 = 5832
We need to find the greatest 10 adjacent numbers

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

The Approach:

We can just iterate through... 1 thousand (minus 10) calculations is ultimately not much for the computer.

To optimize a little better though, we can make assumptions with the window. When we move the window across, if the number going out is larger than the number going in then we don't need to run a calculation... it is going to be smaller than our current max.
Otherwise we will run the calculation.

The Code:

import math

large_number = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
window_size = 13
best_index = 0
best_index_value = math.prod([int(x) for x in large_number[0:window_size]])

for i in range(1,len(large_number) - window_size):
    if int(large_number[i-1]) < int(large_number[i+window_size-1]):
        current_value = math.prod([int(x) for x in large_number[i:i+window_size]])
        if current_value > best_index_value:
            best_index_value = current_value
            best_index = i
            print("newly found at ", best_index, best_index_value)

print(large_number[best_index:best_index+window_size])