# 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

• 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;
}
}