LeetCode 2081: Sum of k-Mirror Numbers Solution

Master LeetCode problem 2081 (Sum of k-Mirror Numbers), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2081. Sum of k-Mirror Numbers

Hard

Math Enumeration

Problem Explanation

Explanation:

To solve this problem, we can iterate through numbers starting from 1 and check if they are k-mirror numbers. We can convert each number to its base-k representation and check if it reads the same both forward and backward. If a number is a k-mirror number, we add it to the sum and continue this process until we find the n smallest k-mirror numbers.

Here is a step-by-step approach:

Initialize a variable to keep track of the count of k-mirror numbers found.
Iterate through numbers starting from 1.
For each number, convert it to base-k representation.
Check if the base-k representation reads the same both forward and backward.
If it does, increment the count and add the number to the sum.
Repeat until n k-mirror numbers are found.

Time Complexity: The time complexity of this approach is O(N * k) where N is the number of k-mirror numbers to find and k is the base.

Space Complexity: The space complexity is O(N) where N is the number of k-mirror numbers to find.

Solution Code

class Solution {
    public int sumOfKMirrorNumbers(int k, int n) {
        int count = 0;
        int sum = 0;
        for (int i = 1; count < n; i++) {
            String numInBaseK = Integer.toString(i, k);
            if (isKMirrorNumber(numInBaseK)) {
                count++;
                sum += i;
            }
        }
        return sum;
    }

    private boolean isKMirrorNumber(String num) {
        return num.equals(new StringBuilder(num).reverse().toString());
    }
}

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