Objective: Given a range of integers, find all the numbers which are palindrome when they are represented in Decimal Value( base 10) and in Octal value(base 8).
Example :
Number : 373 (Decimal) and digits are palindrome. Convert it into Octal which is 565 and that's also palindrome.
Approach:
Solution is quite simple. Traverse through all the numbers in the given range and check if it palindrome, if yes, convert it into Octal and check for palindrome again.
Code:
| public class DecimalOctalPalindrome { | |
| public String DecimalToOctal(int N) { | |
| String Oct = ""; | |
| while (N > 0) { | |
| int x = N % 8; | |
| N = N / 8; | |
| Oct += x; | |
| } | |
| return Oct; | |
| } | |
| public boolean isPalindrome(String S) { | |
| int i = 0; | |
| int j = S.length() – 1; | |
| while (i < j) { | |
| if (S.charAt(i) != S.charAt(j)) { | |
| return false; | |
| } | |
| i++; | |
| j—; | |
| } | |
| return true; | |
| } | |
| public void findBothPalindrome(int start, int end) { | |
| for (int i = start; i <= end; i++) { | |
| String decimal = String.valueOf(i); | |
| if (isPalindrome(decimal)) { | |
| String Oct = DecimalToOctal(i); | |
| if (isPalindrome(Oct)) { | |
| System.out.print(Oct + " "); | |
| } | |
| } | |
| } | |
| } | |
| public static void main(String[] args) { | |
| DecimalOctalPalindrome d = new DecimalOctalPalindrome(); | |
| d.findBothPalindrome(1, 2000); | |
| } | |
| } |
Output:
1 2 3 4 5 6 7 11 171 444 515 565 636 1111