2081. Sum of k-Mirror Numbers
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Problem Overview
Sum of k-Mirror Numbers is a unknown-difficulty LeetCode problem. This is a common Math pattern in coding interviews. Study the solution below and note the time and space complexity before attempting variations on your own.
A full step-by-step explanation is being added. See the study guide for pattern-based practice.
Read the solution code below and trace through it on paper before submitting. For structured interview prep, follow our 30-day study guide.
2081.java
Java
class Solution {
// This method returns the sum of the first n k-mirror numbers.
public long kMirror(int base, int count) {
long sum = 0; // Initialize sum of k-mirror numbers
// Loops indefinitely until 'count' number of k-mirror numbers are found
for (int length = 1;; ++length) {
// Find the start and end range based on the half-length to construct palindrome
// prefixes
int start = (int) Math.pow(10, (length - 1) / 2);
int end = (int) Math.pow(10, (length + 1) / 2);
// Loop to generate all palindrome numbers of the current length
for (int i = start; i < end; i++) {
long palindrome = i;
// Construct the second half of the palindrome number based on the length's
// parity
for (int j = length % 2 == 0 ? i : i / 10; j > 0; j /= 10)
palindrome = palindrome * 10 + j % 10; // Append digits in reverse order to form the palindrome
// Convert palindrome to its representation in the given base
String baseRepresentation = Long.toString(palindrome, base);
// Check if the base representation is a palindrome
if (isPalindrome(baseRepresentation.toCharArray())) {
sum += palindrome; // Add current palindrome to sum
if (--count == 0) // Decrement count and check if 'n' k-mirror numbers are found
return sum; // Return the sum if 'n' k-mirror numbers are found
}
}
}
}
// This method checks if the character array forms a palindrome
private boolean isPalindrome(char[] charArray) {
// Compare characters from opposite ends moving towards the center
for (int i = 0, j = charArray.length - 1; i < j; i++, j--) {
if (charArray[i] != charArray[j])
return false; // Return false if mismatch is found
}
return true; // Return true if the entire array is a palindrome
}
}Was this solution helpful?
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