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.
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.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
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:
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:
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])
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:
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
Case | How to Handle |
---|---|
k is 1 | All 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 0 | Return 0 since we need to sum 0 mirror numbers. |
Integer overflow when calculating the sum | Use 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 k | Base 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 numbers | Implement 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 range | The algorithm should generate enough candidates until n k-mirror numbers are found and summed correctly. |
k-mirror numbers exceeding the maximum integer/long value | Limit 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 range | Dynamically expand the search range for k-mirror numbers until 'n' such numbers are found and ensure the expansion does not lead to overflow. |