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diff --git a/2081. Sum of k-Mirror Numbers/README.md b/2081. Sum of k-Mirror Numbers/README.md
 # 2081. Sum of k-Mirror Numbers
 Hard
 ---
 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. Its representation in base-10 is 9, and in base-2 is 1001, both of which are palindromes.

 - 4 is not a 2-mirror number because its base-2 form is 100, which is not a palindrome.

 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 base-2 representations are:

  base-10    base-2
    1          1
    3         11
    5         101
    7         111
    9        1001

 Sum = 1 + 3 + 5 + 7 + 9 = 25.
 ```

 ### Example 2
 ```
 Input: k = 3, n = 7
 Output: 499
 Explanation:
 The 7 smallest 3-mirror numbers and their base-3 representations are:

  base-10    base-3
    1          1
    2          2
    4         11
    8         22
   121      11111
   151      12121
   212      21212

 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
 ```
diff --git a/2081. Sum of k-Mirror Numbers/Solution.java b/2081. Sum of k-Mirror Numbers/Solution.java
 public class Solution {
    public long kMirror(int k, int n) {
        long ans = 0;
        int cnt = 0;
        long[] pow10 = new long[12];
        pow10[0] = 1;
        for (int i = 1; i < 12; i++) pow10[i] = pow10[i - 1] * 10;

        while (cnt < n) {
            int len = 1;
            for (; ; len++) {
                int half = (len + 1) / 2;
                long start = pow10[half - 1], end = pow10[half];
                for (long h = (half == 1 ? 1 : start); h < end && cnt < n; h++) {
                    long t = h, rev = 0, x = h;
                    if ((len & 1) == 1) t /= 10;
                    while (t > 0) {
                        rev = rev * 10 + t % 10;
                        t /= 10;
                    }
                    long p = x * pow10[len - half] + rev;
                    if (isPal(p, k)) {
                        ans += p;
                        cnt++;
                        if (cnt == n) return ans;
                    }
                }
            }
        }
        return ans;
    }

    private boolean isPal(long x, int k) {
        long t = x, rev = 0;
        while (t > 0) {
            rev = rev * k + t % k;
            t /= k;
        }
        return rev == x;
    }

    // Main function to test the solution
    public static void main(String[] args) {
        Solution sol = new Solution();

        int k1 = 2, n1 = 5;
        System.out.println("Input: k = " + k1 + ", n = " + n1);
        System.out.println("Output: " + sol.kMirror(k1, n1)); // Expected: 25

        int k2 = 3, n2 = 7;
        System.out.println("\nInput: k = " + k2 + ", n = " + n2);
        System.out.println("Output: " + sol.kMirror(k2, n2)); // Expected: 499

        int k3 = 7, n3 = 17;
        System.out.println("\nInput: k = " + k3 + ", n = " + n3);
        System.out.println("Output: " + sol.kMirror(k3, n3)); // Expected: 20379000
    }
 }
Original file line number	Diff line number	Diff line change
		@@ -0,0 +1,62 @@
		# 2081. Sum of k-Mirror Numbers
		Hard
		---
		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. Its representation in base-10 is 9, and in base-2 is 1001, both of which are palindromes.

		- 4 is not a 2-mirror number because its base-2 form is 100, which is not a palindrome.

		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 base-2 representations are:

		base-10 base-2
		1 1
		3 11
		5 101
		7 111
		9 1001

		Sum = 1 + 3 + 5 + 7 + 9 = 25.
		```

		### Example 2
		```
		Input: k = 3, n = 7
		Output: 499
		Explanation:
		The 7 smallest 3-mirror numbers and their base-3 representations are:

		base-10 base-3
		1 1
		2 2
		4 11
		8 22
		121 11111
		151 12121
		212 21212

		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
		```
Original file line number	Diff line number	Diff line change
		@@ -0,0 +1,58 @@
		public class Solution {
		public long kMirror(int k, int n) {
		long ans = 0;
		int cnt = 0;
		long[] pow10 = new long[12];
		pow10[0] = 1;
		for (int i = 1; i < 12; i++) pow10[i] = pow10[i - 1] * 10;

		while (cnt < n) {
		int len = 1;
		for (; ; len++) {
		int half = (len + 1) / 2;
		long start = pow10[half - 1], end = pow10[half];
		for (long h = (half == 1 ? 1 : start); h < end && cnt < n; h++) {
		long t = h, rev = 0, x = h;
		if ((len & 1) == 1) t /= 10;
		while (t > 0) {
		rev = rev * 10 + t % 10;
		t /= 10;
		}
		long p = x * pow10[len - half] + rev;
		if (isPal(p, k)) {
		ans += p;
		cnt++;
		if (cnt == n) return ans;
		}
		}
		}
		}
		return ans;
		}

		private boolean isPal(long x, int k) {
		long t = x, rev = 0;
		while (t > 0) {
		rev = rev * k + t % k;
		t /= k;
		}
		return rev == x;
		}

		// Main function to test the solution
		public static void main(String[] args) {
		Solution sol = new Solution();

		int k1 = 2, n1 = 5;
		System.out.println("Input: k = " + k1 + ", n = " + n1);
		System.out.println("Output: " + sol.kMirror(k1, n1)); // Expected: 25

		int k2 = 3, n2 = 7;
		System.out.println("\nInput: k = " + k2 + ", n = " + n2);
		System.out.println("Output: " + sol.kMirror(k2, n2)); // Expected: 499

		int k3 = 7, n3 = 17;
		System.out.println("\nInput: k = " + k3 + ", n = " + n3);
		System.out.println("Output: " + sol.kMirror(k3, n3)); // Expected: 20379000
		}
		}