2081. Sum of k-Mirror Numbers

Problem Link

Description

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A k-mirror number is a positive integer without leading zeros that reads the same both forward and backward in base-10 as well as in base-k.

• 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

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Python3

class Solution:
    def kMirror(self, k: int, n: int) -> int:
        
        def valid(x):
            res = []
            nums = x
            while x > 0:
                res.append(x % k)
                x //= k
 
            return res == res[::-1]
        
        ans = []
        for d in range(30):
            for i in range(10 ** d, 10 ** (d + 1)):
                x = str(i)
                s = int(x + x[::-1][1:])
                if valid(s):
                    ans.append(s)
                    if len(ans) == n:
                        return sum(ans)
            
            for i in range(10 ** d, 10 ** (d + 1)):
                x = str(i)
                s = int(x + x[::-1])
                if valid(s):
                    ans.append(s)
                    if len(ans) == n:
                        return sum(ans)