Find the character at the `k`th location in the infinite string - c++

Question

I am trying to solve a problem:
Given two strings, s and t, we can form a string x of infinite length, as:
a. Append s to x 1 time;
b. Append t to x 2 times;
c. Append s to x 3 times;
d. Append t to x 4 times;
and so on...
Given k, find the kth character (1 indexed) in the resultant infinite string x.
For e.g., if s = a, t = bc and k = 4, then output: b (x=abcbc). s and t can contain anywhere from 1 to 100 characters, while 1<=k<=10^16.
The brute force way of actually constructing string x is trivial but too slow. How do I optimize it further?
In C++, the brute force solution would look like this:
#include <iostream>
using namespace std;
int main() {
int repeat=1, k=4;
string s="a", t="bc", x;
bool appendS=true;
while(x.size()<k) {
for(int i=1; i<=repeat; i++)
if(appendS) x+=s;
else x+=t;
appendS=!appendS;
repeat++;
}
cout<<x[k-1];
return 0;
}
But how do I optimize it, given huge k?

Answer 1

The string looks like
sttsssttttsssssttttttssssssstttttttt...
Group the string into substrings like
(stts)(ssttttss)(sssttttttsss)(ssssttttttttssss)(sssss...
Let
len(s) = a
len(t) = b
len(s+t) = c
Group 1: stts -> length = 2*c.
Group 2: ssttttss -> length = 4*c.
Group 3: sssttttttsss -> length = 6*c.
Continuing the pattern, it is easy to see that the length of ith group will be 2*i*c.
Let the kth character be in group n.
Total length of first n groups =
2*c + 4*c + 6*c .... + 2*n*c = (2*c)*(1+2+3...+n) = c*n*(n+1)
Since total length of n groups has to be greater than or equal to k,
c*n*(n+1) >= k
n*(n+1) >= k/c
Finding the largest value of n that satisfies this inequality is a trivial task. Now, the nth group looks something like
ss...(n times) + tttt...(2*n times) + ss...(n times)
Now, you just need to find the position of k mod n in this block, which is a simple task.

Answer 2

The search position is located at the nth append, where n is calculated for the last integer where this sum inequality holds, where l1 and l2 are the two strings lengths:
Hence, the search position is at the nth append, at the mod(n+1,2)+1 string, in the Delta k character of that string.
Explanation:
The sum is the addition of all lengths. Since you know all the substrings, you know the whole sum. And since both are integer algebraic expression, you have to do a little linear search over that, to find that integer n, as in any simple integer equation.
Having the nth append number, and the sum, the position is trivially obtained as Delta k.
Moreover, note that Delta k is a closed expression. You do not need any loop to calculate it, just to evaluate, since the sums of the i terms are the sum of the first floor(n/2) even and the first floor(n/2)+1 odd integers.