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Sum of k-Mirror Numbers

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StringsRecursion

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. The representation of 9 in base-10 and base-2 are 9 and 1001 respectively, which read the same both forward and backward.
  • On the contrary, 4 is not a 2-mirror number. The representation of 4 in base-2 is 100, which does not read the same both forward and backward.

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 representations in base-2 are listed as follows:
  base-10    base-2
    1          1
    3          11
    5          101
    7          111
    9          1001
Their sum = 1 + 3 + 5 + 7 + 9 = 25. 

Example 2:

Input: k = 3, n = 7
Output: 499
Explanation:
The 7 smallest 3-mirror numbers are and their representations in base-3 are listed as follows:
  base-10    base-3
    1          1
    2          2
    4          11
    8          22
    121        11111
    151        12121
    212        21212
Their 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

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What are the constraints on `k` and `n`? Specifically, what are the maximum values for `k` and `n`, and can `k` be less than 2?
  2. Are there any constraints on the resulting k-mirror numbers? For instance, is there an upper bound on the magnitude of these numbers that might impact the data type I should use for summation?
  3. If no k-mirror numbers can be found within the first `n` positive integers, what value should I return? Should I return 0, null, or throw an exception?
  4. Are there any specific formatting requirements for the output? For example, should I use a specific data type (e.g., `long`), and should I handle potential overflow issues?
  5. By 'k-mirror number,' do you mean the number is a palindrome when represented in base `k`? For example, if `k` is 2, is 5 (101 in binary) considered a 2-mirror number?

Brute Force Solution

Approach

The brute force approach to this problem involves checking every possible number to see if it meets our requirements. We'll start with the smallest numbers and incrementally check bigger and bigger numbers until we have found enough that meet the criteria.

Here's how the algorithm would work step-by-step:

  1. Start with the number one.
  2. Check if the number is a palindrome in base ten (our normal number system).
  3. Check if the number is also a palindrome in the other base (the 'k' base).
  4. If both are true, then add the number to our collection of valid numbers.
  5. Move on to the next number and repeat steps two through four.
  6. Keep going until we have found the required number of k-mirror numbers (based on the input 'n').
  7. Finally, add all the valid numbers together to get the final sum.

Code Implementation

def sum_of_k_mirror_numbers_brute_force(k_base, number_of_mirror_numbers):

    valid_numbers_found = 0
    current_number = 1
    sum_of_mirror_numbers = 0

    while valid_numbers_found < number_of_mirror_numbers:
        # Check if the number is a palindrome in base 10
        if is_palindrome_base_ten(current_number):

            # Check if the number is also a palindrome in the other base
            if is_palindrome_other_base(current_number, k_base):
                # Accumulate the sum and increment our counter
                sum_of_mirror_numbers += current_number
                valid_numbers_found += 1

        current_number += 1

    return sum_of_mirror_numbers

def is_palindrome_base_ten(number):
    number_string = str(number)
    return number_string == number_string[::-1]

def is_palindrome_other_base(number, k_base):
    converted_number = convert_to_base_k(number, k_base)
    return converted_number == converted_number[::-1]

def convert_to_base_k(number, k_base):
    if number == 0:
        return "0"

    digits = []

    # Repeatedly get the remainders and divide the number
    while number:
        digits.append(str(number % k_base))
        number //= k_base

    # Now reverse the digits to get the result
    return "".join(digits[::-1])

Big(O) Analysis

Time Complexity
O(k * n * log_k(X))The brute force approach iterates through numbers until it finds n k-mirror numbers. Let X be the largest k-mirror number we find. For each number (up to X), we check if it's a palindrome in base 10, which takes O(log X) time, and in base k, which takes O(log_k X) time to convert to base k and O(log_k X) to check if it is a palindrome. Since we're checking up to X numbers and we need to find n k-mirror numbers, the outer loop essentially runs on the order of n times. Thus, the total time complexity is O(X * (log X + k * log_k X)), or roughly O(k * n * log_k(X)), assuming X depends on n and k.
Space Complexity
O(N)The provided brute force approach collects valid k-mirror numbers into a collection. In the worst-case scenario, we might need to store up to N k-mirror numbers, where N is the required number of k-mirror numbers specified in the input. Storing these numbers requires auxiliary space proportional to N. Other temporary variables used for palindrome checks are constant space, so the dominant space complexity is determined by the storage of the k-mirror numbers.

Optimal Solution

Approach

The most efficient way to find k-mirror numbers is to construct them directly instead of checking every number. We generate palindromes in base 10 and base k, then check if both are palindromes. This allows us to find only the numbers we care about.

Here's how the algorithm would work step-by-step:

  1. Start by generating potential mirror numbers, building them from the center outwards, both for odd and even lengths.
  2. For each potential mirror number, check if it's a palindrome when written in base 10.
  3. If it is a palindrome in base 10, convert the number to base k.
  4. Check if the base k representation is also a palindrome.
  5. If both base 10 and base k representations are palindromes, add the number to our sum.
  6. Repeat this process, generating larger and larger potential mirror numbers, until we've found enough of them to meet the problem's requirements.

Code Implementation

def sum_k_mirror_numbers(k_base, number_of_terms):
    def is_palindrome(number_string):
        return number_string == number_string[::-1]

    def convert_to_base_k(number, k_base):
        if number == 0:
            return '0'
        digits = []
        while number:
            digits.append(str(number % k_base))
            number //= k_base
        return ''.join(digits[::-1])

    def generate_palindromes():
        # Start with single digit palindromes
        for i in range(1, 10):
            yield str(i)

        length = 2
        while True:
            # Generate odd length palindromes
            for middle_digit in range(10):
                half = str(10**(length // 2))
                start = int(half)
                end = int(str(10**(length // 2 +1) -1))
                for i in range(start,end+1):
                    string_num = str(i)
                    palindrome = string_num + str(middle_digit) + string_num[::-1]
                    yield palindrome

            # Generate even length palindromes
            half = str(10**(length // 2-1))
            start = int(half)
            end = int(str(10**(length // 2) -1))
            for i in range(start,end+1):
                string_num = str(i)
                palindrome = string_num + string_num[::-1]
                yield palindrome
            length += 1

    sum_of_mirror_numbers = 0
    count = 0

    #Iterate through generator
    for potential_palindrome in generate_palindromes():
        decimal_representation = int(potential_palindrome)
        #Only analyze when sufficient terms have not been reached
        if count < number_of_terms:
            if is_palindrome(str(decimal_representation)):
                base_k_representation = convert_to_base_k(decimal_representation, k_base)

                #Ensure the base K is a palindrome
                if is_palindrome(base_k_representation):
                    sum_of_mirror_numbers += decimal_representation
                    count += 1
        else:
            break
    
    return sum_of_mirror_numbers

Big(O) Analysis

Time Complexity
O(n * log_k(n))The algorithm generates potential mirror numbers until n such numbers are found. Generating each number involves creating base-10 palindromes, which takes time proportional to the number of digits. Then each base-10 palindrome is converted to base k. Converting a base-10 number n to base k takes O(log_k(n)) time. Since we are summing k-mirror numbers until we reach n such numbers, where n is not the input, but the number of k-mirror numbers we are finding, generating each candidate and verifying it takes O(log_k(candidate_number)) for base conversion. Since we are generating numbers to find n k-mirror numbers, we can say we are doing base k conversions log_k(n) times for each of the n k-mirror numbers found. So overall, the time complexity is O(n * log_k(n)).
Space Complexity
O(D)The dominant space complexity factor arises from generating potential mirror numbers and converting them to base k. While generating palindromes could lead to arbitrarily large numbers, we stop when we've found enough mirror numbers. Let D be the maximum number of digits in base k for a k-mirror number encountered. The space used to store the number in base k as a string or an array of digits is therefore O(D). Thus, the auxiliary space complexity is O(D), where D is the maximum number of digits in base k of the k-mirror numbers found.

Edge Cases

CaseHow to Handle
k is 1All numbers are base-1 palindromes, so the generation and filtering process needs to be adapted or short-circuited to avoid infinite loop, potentially returning first n numbers.
n is 0Return 0 since we need to sum 0 mirror numbers.
Integer overflow when calculating the sumUse a data type with a larger range (e.g., long) or handle overflow explicitly with modular arithmetic, if applicable, depending on problem contraints (return modulo a certain number)
Large values of kBase conversion might become computationally expensive, thus optimization may be required or the problem itself may only be solvable for small values of k.
n is very large, requiring generation of many k-mirror numbersImplement an efficient mirror number generation algorithm that avoids unnecessary computations and potential time out.
Small values of k (e.g., k=2) leading to fewer k-mirror numbers in the desired rangeThe algorithm should generate enough candidates until n k-mirror numbers are found and summed correctly.
k-mirror numbers exceeding the maximum integer/long valueLimit the range of generated mirror numbers based on the maximum representable value and adjust the number generation strategy accordingly to avoid overflow errors.
Cases where no k-mirror numbers exist within the initially tested rangeDynamically expand the search range for k-mirror numbers until 'n' such numbers are found and ensure the expansion does not lead to overflow.