Block sequence

• Time : $O(log(n))$
• Space : $O(1)$

int solve(long long n){
    for(long long i = 1, sum = 1, add = 1, po = 10; n > sum;) {
        n -= sum;
        i += 1;
        if(i == po) {
            add += 1;
            po *= 10;
        }
        sum += add;
    }
    for(long long i = 1, add = 1, po = 10;;) {
        if(n > add) n -= add;
        else {
            return std::to_string(i)[n-1] - '0';
        }
        i += 1;
        if(i == po) {
            add += 1, po *= 10;
        }
    }
    return -1;
}