Unlucky number 13

I think this can be solved via recursion:

ans(n) = { ans([n/2])^2 - ans([n/2]-1)^2 }, if n is even
ans(n) = { ans([n/2]+1)*ans([n/2]) - ans([n/2])*ans([n/2]-1) }, if n is odd

Base Cases:

• ans(0) = 1
• ans(1) = 10

It's implementation is running quite fast even for larger inputs like 10^9 ( which is expected as its complexity is O(log[n]) instead of O(n) like the other answers ):

cache = {}

mod = 1000000009

def ans(n):
    if cache.has_key(n):
        return cache[n]

    if n == 0:
        cache[n] = 1
        return cache[n]
    if n == 1:
        cache[n] = 10
        return cache[n]

    temp1 = ans(n/2)
    temp2 = ans(n/2-1)

    if (n & 1) == 0:
        cache[n] = (temp1*temp1 - temp2*temp2) % mod
    else:
        temp3 = ans(n/2 + 1)
        cache[n] = (temp1 * (temp3 - temp2)) % mod

    return cache[n]

print ans(1000000000)

Online Demo

Explanation:

Let a string s have even number of digits 'n'.
Let ans(n) be the answer for the input n, i.e. the number of strings without the substring 13 in them.
Therefore, the answer for string s having length n can be written as the multiplication of the answer for the first half of the string (ans([n/2])) and the answer for the second half of the string (ans([n/2])), minus the number of cases where the string 13 appears in the middle of the number n, i.e. when the last digit of the first half is 1 and the first digit of the second half is 3.

This can expressed mathematically as:

ans(n) = ans([n/2])^2 - ans([n/2]-1)*2

Similarly for the cases where the input number n is odd, we can derive the following equation:

ans(n) = ans([n/2]+1)*ans([n/2]) - ans([n/2])*ans([n/2]-1)

I get the feeling that this question is designed with the expectation that you would initially instinctively do it the way you have. However, I believe there's a slightly different approach that would be faster.

You can produce all the numbers that contain the number 13 yourself, without having to loop through all the numbers in between. For example:

2 digits: 13

3 digits position 1: 113 213 313 etc.

3 digits position 2: 131 132 133 etc.

Therefore, you don't have to check all the number from 0 to n*9. You simply count all the numbers with 13 in them until the length is larger than N.

This may not be the fastest solution (in fact I'd be surprised if this couldn't be solved efficiently by using some mathematics trickery) but I believe it will be more efficient than the approach you have currently taken.