2081. Sum of k-Mirror Numbers LeetCode Solution

Last updated: February 26, 2025

In this guide, you will get 2081. Sum of k-Mirror Numbers LeetCode Solution with the best time and space complexity. The solution to Sum of k-Mirror Numbers problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Sum of k-Mirror Numbers solution in C++
Sum of k-Mirror Numbers solution in Java
Sum of k-Mirror Numbers solution in Python
Additional Resources

2081. Sum of k-Mirror Numbers LeetCode Solution image

Problem Statement of Sum of k-Mirror Numbers

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

Complexity Analysis

Time Complexity:
Space Complexity:

2081. Sum of k-Mirror Numbers LeetCode Solution in C++

class Solution:
  def kMirror(self, k: int, n: int) -> int:
    ans = 0
    A = ['0']

    def nextKMirror(A: list[str]) -> list[str]:
      for i in range(len(A) // 2, len(A)):
        nextNum = int(A[i]) + 1
        if nextNum < k:
          A[i] = str(nextNum)
          A[~i] = str(nextNum)
          for j in range(len(A) // 2, i):
            A[j] = '0'
            A[~j] = '0'
          return A
      return ['1'] + ['0'] * (len(A) - 1) + ['1']

    for _ in range(n):
      while True:
        A = nextKMirror(A)
        num = int(''.join(A), k)
        if str(num)[::-1] == str(num):
          break
      ans += num

    return ans
/* code provided by PROGIEZ */

2081. Sum of k-Mirror Numbers LeetCode Solution in Java

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2081. Sum of k-Mirror Numbers LeetCode Solution in Python

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Additional Resources

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