Problem Description

A k-mirror number is a positive integer with no leading zeros that is symmetrical; it reads the same forward and backward. This symmetry must hold true when the number is expressed in both base-10 (our standard numeral system) and base-k (a numeral system with base k). For example, the number 9 is the same whether read forwards or backwards in base-10, and when converted to base-2, it becomes '1001', which is also symmetrical.

However, not every number is k-mirror. For instance, the number 4 in base-10 becomes '100' in base-2. Since '100' isn't symmetrical, 4 is not a 2-mirror number.

The problem asks us to calculate the sum of the n smallest k-mirror numbers. In other words, we must find the first n numbers that are k-mirror numbers and add them together.

Intuition

The solution approach for finding k-mirror numbers involves generating potential mirror numbers in base-10 and then checking if they are also symmetrical in base-k. Instead of generating all numbers and checking each one, a more efficient approach is to generate numbers that are already palindromic in base-10. If a number is a palindrome in base-10, there's no need to check it for symmetry in this base.

To generate palindromes in base-10, we start by considering numbers of increasing lengths. For each length, we form the first half of a potential palindrome and then mirror it to create the second half, ensuring the entire number is symmetrical in base-10.

For example, if our first half is '123', the whole palindrome can be '12321' for odd lengths, and '123321' for even lengths. Note that we have to generate both even and odd length palindromes because they may both yield valid k-mirror numbers.

Once we create a base-10 palindrome, we convert it to base-k and then check if the base-k representation is also a palindrome. The check is straightforward: compare the first half of the array of characters to the second half and confirm they are identical.

The provided Java function kMirror(int k, int n) initializes the answer to 0. It then enters a loop that continues until we find n k-mirror numbers. Inside the loop, we generate base-10 palindromes of increasing length using the loop variable l. We use the just-generated palindrome as a long integer v, convert v to a string in base-k, and then use the check helper method to determine if it is also a palindrome in base-k.

Each time we find a k-mirror number, we add it to the ans variable and decrement n. Once we've found n k-mirror numbers, we return the total sum ans.

This approach is efficient because it sidesteps the need to check each base-10 number for palindromic properties—we create only numbers that are already palindromes in base-10. Then we only need to check for palindromic structure in base-k, significantly reducing the number of checks needed.

Learn more about Math patterns.

Solution Approach

The solution uses a mathematical approach to generate palindromes and a brute-force check to determine if those palindromes are k-mirror numbers. Let's break down the implementation into its main components:

1. Palindrome Generation: The outer loop, indicated by for (int l = 1;; ++l), controls the length l of the palindromes being generated. The l increases after each full iteration, allowing us to create larger palindromes. For half the length (rounded down), we use int x = (int) [Math](/problems/math-basics).pow(10, (l - 1) / 2), and for half the length (rounded up), we use int y = (int) Math.pow(10, (l + 1) / 2).

2. Base-10 Palindrome Construction: Inside the loop, for (int i = x; i < y; i++) iterates through potential first halves of the palindromic numbers. The variable v is then used to construct the full palindrome by appending the reverse of i, either including the last digit (l % 2 == 0 ? i : i / 10) for even lengths or excluding it for odd lengths.

3. Base-k Conversion: After creating a base-10 palindrome (v), we convert it to a string in the base-k numeral system with String ss = Long.toString(v, k).

4. Palindrome Check in Base-k: The private helper function check(char[] c) takes in the char array of the base-k representation and uses two pointers (i and j) to check from both ends towards the center, ensuring that the string is the same forwards and backwards.

5. Summation and Counting: When a k-mirror number is found (i.e., it passes the palindrome check in base-k), it is added to the running total ans, and n is decremented. We continue generating palindromes and checking for k-mirror symmetry until n reaches zero—at which point we return the sum ans.

The algorithm demonstrates efficient use of mathematical properties to limit the search space only to valid palindromes, rather than iterating through all integers. By constructing palindromes of increasing length, we also ensure we are considering the smallest k-mirror numbers in ascending order, which fits the requirement of the problem. The combination of mathematical pattern recognition and brute-force checking results in a direct and effective solution to the problem.

Example Walkthrough

Let's use k = 2 and n = 3 to illustrate the solution approach to finding the sum of the n smallest k-mirror numbers. We are looking for the sum of the first 3 numbers that are both palindromes in base-10 and base-2.

1. Palindrome Generation in Base-10: Our outer loop constructs base-10 palindromes of increasing length. We begin with the length l = 1. This means for half the length (rounded down) x=1 and half the length (rounded up) y=10.

2. Base-10 Palindrome Construction: Inside the loop, we iterate from x (in this case, 1) to y (less than 10) to create the first half of the palindrome. For i = 1, we get v = 1, because this is a length one (odd-length) palindrome i itself. For an even-length palindrome when l = 2, which is not applicable yet, we would mirror the whole digit(s).

3. Base-k Conversion: Now, we convert v to base-k (base-2). The number 1 in base-2 is '1', which is still a palindrome.

4. Palindrome Check in Base-k: For our number 1 in base-2, it's clearly symmetrical, so no need to check further.

5. Summation and Counting: Since 1 is a k-mirror number, we add it to the running total ans = 1 and decrement n to 2. We now need to find two more k-mirror numbers.

Next, for i = 2, repeating the steps above, we:

1. Create the base-10 palindrome, which is still 2 since it's a single digit.
2. Convert 2 into base-2, which is '10', and this is not a palindrome. Hence, we skip it.

For i = 3, we:

1. Have v = 3, already a palindrome in base-10.
2. Convert 3 into base-2, which gives us '11'. It is a palindrome in base-2.
3. Since '11' passes the symmetry check, we add 3 to ans, resulting in ans = 4, and decrement n to 1.

For l greater than 1, let's say l = 2, our process works as follows:

1. Generate potential base-10 palindromes by considering i from 1 to 9 (the upper limit y here would be 10).
2. For i = 1, our even base-10 palindrome construction would yield 11.
3. Convert 11 into base-2, which gives us '1011', and this is not symmetrical.
4. Thus, 11 in base-10 is not added to ans and won't decrement n.

This process repeats, increasing l as needed, until we find 3 k-mirror numbers in base-10 and base-2. Once we find the required amount, we return the total sum ans, which would be the sum of the first three smallest 2-mirror numbers.

The efficiency of this algorithm comes from the fact that rather than brute-forcing through consecutive integers, it constructs potential palindromes and then checks if those palindromes are symmetrical in another base, significantly narrowing the search space.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code can be broken down as follows:

1. The outermost loop runs indefinitely (for (int l = 1;; ++l)), incrementing l until n palindromes are found. The value l determines the length of the palindrome to be constructed.

2. Inside the outermost for loop, calculations to set the bounds x and y are done using Math.pow(10, (l - 1) / 2) and Math.pow(10, (l + 1) / 2), which has a constant time complexity of O(1) as it is executed once per loop iteration.

3. The next for loop (for (int i = x; i < y; i++)) iterates from x to y. The number of iterations depends on l and is approximately O(10^(l/2)).

4. Inside this loop, the palindrome is constructed by appending the reverse of the number to itself or its prefix depending on whether l is even or odd. This operation has a time complexity of O(l) per iteration since reversing a number's digits is linear with the number of digits.

5. The palindrome is then converted to its base-k representation (Long.toString(v, k)). The conversion's time complexity is O((log(v)/log(k)) * log(v)), where (log(v)/log(k)) is the length of the string in base-k.

6. The check function, which ensures the base-k string is a palindrome, has a time complexity of O(log(v)/log(k)).

Considering that v grows exponentially with l, the dominant factor here is the construction and checking of base-k palindromes which happen (10^{l/2}) times for each length l. Thus, the total time complexity approximately can be expressed as:

O(n * l * (log(v)/log(k)) * log(v)), as this represents the work done to find each of the n palindromes.

Given that n is the number of palindromes required and l increases incrementally, this expression accounts for the increasing cost of producing and checking each palindrome as their lengths increase.

Space Complexity

The space complexity of the code can be analyzed as follows:

1. The variable v and the string ss carry the largest values in this code. Since these are reused for each palindrome, they do not increase in space usage proportional to the input size.

2. The check function uses a few variables for iteration, which does not significantly contribute to the space complexity.

3. No additional arrays or data structures are used that grow with input size.

Thus, the space complexity is primarily driven by the call stack for each recursive call used in number conversion and checking, which yields:

• O(l) for the recursion in creating the palindrome.
• O(log(v)) for space required for base-k conversion.

Since l is linearly related to log(v) (as the palindrome v will have a length corresponding to l), the space complexity is approximately O(log(v)). However, because log(v) is a constant relative to v itself and does not grow with the input size n, the overall space complexity of the code is O(1), i.e., constant space complexity assuming that k is also constant.

Learn more about how to find time and space complexity quickly using problem constraints.