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2081. Sum of k-Mirror Numbers

Description

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

Solutions

Solution 1

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class Solution {
    public long kMirror(int k, int n) {
        long ans = 0;
        for (int l = 1;; ++l) {
            int x = (int) Math.pow(10, (l - 1) / 2);
            int y = (int) Math.pow(10, (l + 1) / 2);
            for (int i = x; i < y; i++) {
                long v = i;
                for (int j = l % 2 == 0 ? i : i / 10; j > 0; j /= 10) {
                    v = v * 10 + j % 10;
                }
                String ss = Long.toString(v, k);
                if (check(ss.toCharArray())) {
                    ans += v;
                    if (--n == 0) {
                        return ans;
                    }
                }
            }
        }
    }

    private boolean check(char[] c) {
        for (int i = 0, j = c.length - 1; i < j; i++, j--) {
            if (c[i] != c[j]) {
                return false;
            }
        }
        return true;
    }
}

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