This content originally appeared on DEV Community and was authored by Berlin Techs
👋 Introduction
In the world of programming puzzles, palindrome-based problems always offer a unique challenge—especially when they span across multiple number bases. Imagine trying to find numbers that not only look the same forwards and backwards in base-10, but also in base-k. That’s the crux of the k-mirror number challenge.
Let’s dive into the formal problem definition to see how this works:
🧠 Problem Summary
You're given two integers:
k: the base in which we check for palindromes
n: the number of k-mirror numbers to find
A k-mirror number is a positive integer that:
Is a palindrome in base-10
Is also a palindrome in base-k
Your task: return the sum of the first n such numbers.
🧩 Intuition
This is a palindrome-generation and filtering problem. The challenge lies in generating candidate numbers efficiently and checking their representations in different bases.
Key Observations:
Not every decimal palindrome is a k-palindrome (i.e., palindrome in base-k)
Instead of checking all numbers, generate only palindromes in base-10 (which drastically reduces search space)
For each generated number, convert it to base-k and check if that string is also a palindrome
🪄 Approach
Start from the smallest base-10 palindromes: 1-digit numbers.
Use a helper function to generate the next decimal palindrome.
For each candidate:
Convert it to base-k
Check if that base-k representation is a palindrome
If it is, add it to the sum and count
Stop when we find n such numbers
💻 C++ Code
class Solution {
public:
bool isPalindrome(string s) {
return s == string(s.rbegin(), s.rend());
}
string toBaseK(long long num, int k) {
string res;
while (num > 0) {
res += to_string(num % k);
num /= k;
}
reverse(res.begin(), res.end());
return res;
}
long long kMirror(int k, int n) {
long long sum = 0;
int len = 1;
while (n > 0) {
for (int half = pow(10, (len - 1) / 2); half < pow(10, (len + 1) / 2) && n > 0; ++half) {
string h = to_string(half);
string r = h;
reverse(r.begin(), r.end());
string full = h + (len % 2 ? r.substr(1) : r);
long long num = stoll(full);
if (isPalindrome(toBaseK(num, k))) {
sum += num;
--n;
}
}
++len;
}
return sum;
}
};
💻 JavaScript Code
function isPalindrome(s) {
return s === s.split('').reverse().join('');
}
function toBaseK(num, k) {
return num.toString(k);
}
var kMirror = function(k, n) {
let sum = 0;
let len = 1;
while (n > 0) {
let lower = Math.pow(10, Math.floor((len - 1) / 2));
let upper = Math.pow(10, Math.ceil((len) / 2));
for (let half = lower; half < upper && n > 0; half++) {
let h = String(half);
let r = h.split('').reverse().join('');
let full = h + (len % 2 ? r.slice(1) : r);
let num = parseInt(full);
if (isPalindrome(toBaseK(num, k))) {
sum += num;
n--;
}
}
len++;
}
return sum;
};
🐍 Python Code
class Solution:
def kMirror(self, k: int, n: int) -> int:
def is_palindrome(s):
return s == s[::-1]
def to_base_k(num, k):
res = ""
while num:
res = str(num % k) + res
num //= k
return res
def generate_palindromes():
length = 1
while True:
for half in range(10 ** ((length - 1) // 2), 10 ** ((length + 1) // 2)):
s = str(half)
yield int(s + s[-2::-1] if length % 2 else s + s[::-1])
length += 1
ans = 0
count = 0
for num in generate_palindromes():
if is_palindrome(to_base_k(num, k)):
ans += num
count += 1
if count == n:
break
return ans
This content originally appeared on DEV Community and was authored by Berlin Techs
Berlin Techs | Sciencx (2025-06-27T05:13:41+00:00) 🏂 Beginner-Friendly Guide: “Sum of k-Mirror Numbers” – LeetCode 2081 (python| JavaScript | Python ). Retrieved from https://www.scien.cx/2025/06/27/%f0%9f%8f%82-beginner-friendly-guide-sum-of-k-mirror-numbers-leetcode-2081-python-javascript-python/
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