How do you solve Sum of k-Mirror Numbers on LeetCode?

Sum of k-Mirror Numbers (LeetCode #2081) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Generating palindromic numbers directly reduces the search space significantly.

What is the brute force solution for Sum of k-Mirror Numbers?

The brute force approach for Sum of k-Mirror Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each number sequentially to see if it is a k-mirror number. This involves checking if the number is a palindrome in both base-10 and base-k, which can be slow but is straightforward. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of k-Mirror Numbers?

Sum of k-Mirror Numbers uses the following concepts: Math, Enumeration. The recommended patterns to study are: Enumeration, Backtracking (for generating palindromes).

What are the interview tips for Sum of k-Mirror Numbers?

Always clarify the constraints and edge cases before diving into coding. Think about how to reduce the problem size — generating valid inputs can save time. Practice converting between bases and checking for palindromes as these are common tasks.

What are common mistakes when solving Sum of k-Mirror Numbers?

Not checking for leading zeros when generating palindromes. Overlooking the efficiency of generating palindromes instead of checking each number.

#2081

Sum of k-Mirror Numbers

Hard

MathEnumerationEnumerationBacktracking (for generating palindromes)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking every number, we can generate palindromic numbers directly. This reduces the search space significantly and allows us to find k-mirror numbers more efficiently.

⚙️

Algorithm

3 steps

1Step 1: Generate palindromic numbers by constructing them from half the digits.
2Step 2: For each generated palindrome, check if it is a palindrome in base-k.
3Step 3: Keep track of the count and sum of valid k-mirror numbers until we reach 'n'.

solution.py30 lines

1# Full working Python code
2
3def generate_palindromes(max_digits):
4    palindromes = []
5    for length in range(1, max_digits + 1):
6        half_length = (length + 1) // 2
7        for i in range(10 ** (half_length - 1), 10 ** half_length):
8            half = str(i)
9            if length % 2 == 0:
10                palindromes.append(int(half + half[::-1]))
11            else:
12                palindromes.append(int(half + half[-2::-1]))
13    return palindromes
14
15
16def k_mirror(k, n):
17    palindromes = generate_palindromes(7)  # Generate palindromes with up to 7 digits
18    count = 0
19    sum_mirror = 0
20    for num in palindromes:
21        if count >= n:
22            break
23        base_k_repr = to_base_k(num, k)
24        if is_palindrome(base_k_repr):
25            sum_mirror += num
26            count += 1
27    return sum_mirror
28
29# Example usage
30print(k_mirror(2, 5))  # Output: 25

ℹ

Complexity note: The time complexity is O(n) because we generate palindromic numbers directly, significantly reducing the number of checks needed compared to the brute force approach.

1Generating palindromic numbers directly reduces the search space significantly.
2Understanding how to convert numbers between bases is crucial for this problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.