Double Base Palindrome

Double base Palindrome as the name suggest is a number which is Palindrome in 2 bases. One of the base is 10 i.e. decimal and another base is k.(which can be 2 or others). 
Note : The palindromic number, in either base, may not include leading zeros. 
Example : The decimal number, 585 = 1001001001₂ (binary), is palindromic in both bases. 

A Palindrome is a word, phrase, number, or other sequence of characters which reads the same backward as forward, such as madam or 12321.

Find the sum of all numbers less than n which are palindromic in base 10 and base k.

Examples: 

Input : 10 2
Output : 25
Explanation : (here n = 10 and k = 2)
 1 3 5 7 9 (they are all palindrome 
 in base 10 and 2) so sum is :
 1 + 3 + 5 + 7 + 9 = 25

Input : 100 2
Output : 157
Explanation : 1 + 3 + 5 + 7 + 9 + 33 + 99 = 157

Method 1 : This method is simple. For every number less than n : 

• Check if it is a palindrome in base 10
• If yes, then convert it into base k
• Check if it is palindrome in base k
• If yes, then add it in sum.

This method is quite lengthy as it checks for every number whether it is a palindrome or not. So, for number as large as 1000000, it checks for every number. 
If k = 2, then a palindrome in base 2 can only be odd number, which might reduce the comparisons to 1000000 / 2 = 500000 (which is still large).

Below is the implementation of the above approach :  

Total sum is 157

Time Complexity: O(n*log(n))
Auxiliary Space: O(log(n))

Method 2 : This method is a little complex to understand but more advance than method 1. Rather than checking palindrome for two bases. This method generates palindrome in given range. 
Suppose we have a palindrome of the form 123321 in base k, then the first 3 digits define the palindrome. However, the 3 digits 123 also define the palindrome 12321. So the 3-digit number 123 defines a 5-digit palindrome and a 6 digit palindrome. From which follows that every positive number less than kⁿ generates two palindromes less than k²ⁿ . This holds for every base k. Example : let's say k = 10 that is decimal. Then for n = 1, all numbers less than 10ⁿ have 2 palindrome, 1 even length and 1 odd length in 10²ⁿ. These are 1, 11 or 2, 22 or 3, 33 and so on. So, for 1000000 we generate around 2000 and for 10⁸ we generate around 20000 palindromes. 

• Start from i=1 and generate odd palindrome of it.
• Check if this generated odd palindrome is also palindrome in base k
• If yes, then add this number to sum.
• Repeat the above three steps by changing i=i+1 until the last generated odd palindrome has crossed limit.
• Now, again start from i=1 and generate even palindrome of it.
• Check if this generated even palindrome is also palindrome in base k
• If yes, then add this number to sum.
• Repeat the above three steps by changing i=i+1 until the last generated even palindrome has crossed limit.

Below is the implementation of the above approach :  

Total sum is 872187

Time Complexity: O(n*log(n))
Auxiliary Space: O(1)