• For example, 9 is a 2-mirror number. The representation of 9 in base-10 and base-2 are 9 and 1001 respectively, which read the same both forward and backward.
• On the contrary, 4 is not a 2-mirror number. The representation of 4 in base-2 is 100, which does not read the same both forward and backward.

Given the base k and the number n, return the sum of the n smallest k-mirror numbers.

Example 1:

Input: k = 2, n = 5
Output: 25
Explanation:
The 5 smallest 2-mirror numbers and their representations in base-2 are listed as follows:
  base-10    base-2
    1          1
    3          11
    5          101
    7          111
    9          1001
Their sum = 1 + 3 + 5 + 7 + 9 = 25. 

Example 2:

Input: k = 3, n = 7
Output: 499
Explanation:
The 7 smallest 3-mirror numbers are and their representations in base-3 are listed as follows:
  base-10    base-3
    1          1
    2          2
    4          11
    8          22
    121        11111
    151        12121
    212        21212
Their sum = 1 + 2 + 4 + 8 + 121 + 151 + 212 = 499.

Example 3:

Input: k = 7, n = 17
Output: 20379000
Explanation: The 17 smallest 7-mirror numbers are:
1, 2, 3, 4, 5, 6, 8, 121, 171, 242, 292, 16561, 65656, 2137312, 4602064, 6597956, 6958596

Constraints:

• 2 <= k <= 9
• 1 <= n <= 30

Solution: Generate palindromes in base-k.

# Author: Huahua
class Solution:
  def kMirror(self, k: int, n: int) -> int:
    def getNext(x: str) -> str:
      s = list(x)
      l = len(s)    
      for i in range(l // 2, l):
        if int(s[i]) + 1 >= k: continue
        s[i] = s[~i] = str(int(s[i]) + 1)
        for j in range(l // 2, i): s[j] = s[~j] = "0"
        return "".join(s)
      return "1" + "0" * (l - 1) + "1"
    ans = 0
    x = "0"
    for _ in range(n):
      while True:
        x = getNext(x)
        val = int(x, k)
        if str(val) == str(val)[::-1]: break
      ans += val
    return ans

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