Find the number at the n position in the infinite sequence - swift

Question

Having an infinite sequence s = 1234567891011...
Let's find the number at the n position (n <= 10^18)
EX: n = 12 => 1; n = 15 => 2
import Foundation
func findNumber(n: Int) -> Character {
var i = 1
var z = ""
while i < n + 1 {
z.append(String(i))
i += 1
}
print(z)
return z[z.index(z.startIndex, offsetBy: n-1)]
}
print(findNumber(n: 12))
That's my code but when I find the number at 100.000th position, it returns an error, I thought I appended too many i to z string.
Can anyone help me, in swift language?

Answer 1

The problem we have here looks fairly straight forward. Take a list of all the number 1-infinity and concatenate them into a string. Then find the nth digit. Straight forward problem to understand. The issue that you are seeing though is that we do not have an infinite amount of memory nor time to be able to do this reasonably in a computer program. So we must find an alternative way around this that does not just add the numbers onto a string and then find the nth digit.
The first thing we can say is that we know what the entire list is. It will always be the same. So can we use any properties of this list to help us?
Let's call the input number n. This is the position of the digit that we want to find. Let's call the output digit d.
Well, first off, let's look at some examples.
We know all the single digit numbers are just in the same position as the number itself.
So, for n<10 ... d = n
What about for two digit numbers?
Well, we know that 10 starts at position 10. (Because there are 9 single digit numbers before it). 9 + 1 = 10
11 starts at position 12. Again, 9 single digits + one 2 digit number before it. 9 + 2 + 1 = 12
So how about, say... 25? Well that has 9 single digit numbers and 15 two digit numbers before it. So 25 starts at 9*1 + 15*2 + 1 = 40 (+ 1 as the sum gets us to the end of 24 not the start of 25).
So... 99 starts at? 9*1 + 89*2 + 1 = 188.
The we do the same for the three digit numbers...
100... 9*1 + 90*2 + 1 = 190
300... 9*1 + 90*2 + 199*3 + 1 = 787
1000...? 9*1 + 90*2 + 900*3 + 1 = 2890
OK... so now I'm seeing a pattern here that seems to need to know the number of digits in each number. Well... I can get the number of digits in a number by rounding up the log(base 10) of that number.
rounding up log base 10 of 5 = 1
rounding up log base 10 of 23 = 2
rounding up log base 10 of 99 = 2
rounding up log base 10 of 627 = 3
OK... so I think I need something like...
// in pseudo code
let lengthOfNumber = getLengthOfNumber(n)
var result = 0
for each i from 0 to lengthOfNumber - 1 {
result += 9 * 10^i * (i + 1) // this give 9*1 + 90*2 + 900*3 + ...
}
let remainder = n - 10^(lengthOfNumber - 1) // these have not been added in the loop above
result += remainder * lengthOfNumber
So, in the above pseudo code you can give it any number and it will return the position in the list that that number starts on.
This isn't the exact same as the problem you are trying to solve. And I don't want to solve it for you.
This is just a leg up on how I would go about solving it. Hopefully, this will give you some guidance on how you can take this further and solve the problem that you are trying to solve.