JavaAlgorithms/Misc/LowestBasePalindrome.java

import java.util.InputMismatchException;
import java.util.Scanner;

/**
 * Class for finding the lowest base in which a given integer is a palindrome.
 * Includes auxiliary methods for converting between bases and reversing strings.
 * 
 * NOTE: There is potential for error, see note at line 63.
 * 
 * @author RollandMichael
 * @version 2017.09.28
 *
 */
public class LowestBasePalindrome {
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n=0;
		while (true) {
			try {
				System.out.print("Enter number: ");
				n = in.nextInt();
				break;
			} catch (InputMismatchException e) {
				System.out.println("Invalid input!");
				in.next();
			}
		}
		System.out.println(n+" is a palindrome in base "+lowestBasePalindrome(n));
		System.out.println(base2base(Integer.toString(n),10, lowestBasePalindrome(n)));
	}
	
	/**
	 * Given a number in base 10, returns the lowest base in which the
	 * number is represented by a palindrome (read the same left-to-right
	 * and right-to-left).
	 * @param num A number in base 10.
	 * @return The lowest base in which num is a palindrome.
	 */
	public static int lowestBasePalindrome(int num) {
		int base, num2=num;
		int digit;
		char digitC;
		boolean foundBase=false;
		String newNum = "";
		String digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
		
		while (!foundBase) {
			// Try from bases 2 to num (any number n in base n is 1)
			for (base=2; base<num2; base++) {
				newNum="";
				while(num>0) {
					// Obtain the first digit of n in the current base,
					// which is equivalent to the integer remainder of (n/base).
					// The next digit is obtained by dividing n by the base and
					// continuing the process of getting the remainder. This is done
					// until n is <=0 and the number in the new base is obtained.
					digit = (num % base);
					num/=base;
					// If the digit isn't in the set of [0-9][A-Z] (beyond base 36), its character
					// form is just its value in ASCII.
					
					// NOTE: This may cause problems, as the capital letters are ASCII values
					// 65-90. It may cause false positives when one digit is, for instance 10 and assigned
					// 'A' from the character array and the other is 65 and also assigned 'A'.
					
					// Regardless, the character is added to the representation of n
					// in the current base.
					if (digit>=digits.length()) {
						digitC=(char)(digit);
						newNum+=digitC;
						continue;
					}
					newNum+=digits.charAt(digit);
				}
				// Num is assigned back its original value for the next iteration.
				num=num2;
				// Auxiliary method reverses the number.
				String reverse = reverse(newNum);
				// If the number is read the same as its reverse, then it is a palindrome.
				// The current base is returned.
				if (reverse.equals(newNum)) {
					foundBase=true;
					return base;
				}
			}
		}
		// If all else fails, n is always a palindrome in base n-1. ("11")
		return num-1;
	}
	
	private static String reverse(String str) {
	    String reverse = "";
	    for(int i=str.length()-1; i>=0; i--) {
	       reverse += str.charAt(i);
	    }
	    return reverse;
	}
	
	private static String base2base(String n, int b1, int b2) {
		// Declare variables: decimal value of n, 
		// character of base b1, character of base b2,
		// and the string that will be returned.
		int decimalValue = 0, charB2;
		char charB1;
		String output="";
		// Go through every character of n
		for (int i=0; i<n.length(); i++) {
			// store the character in charB1
			charB1 = n.charAt(i);
			// if it is a non-number, convert it to a decimal value >9 and store it in charB2
			if (charB1 >= 'A' && charB1 <= 'Z') 
				charB2 = 10 + (charB1 - 'A');
			// Else, store the integer value in charB2
			else 
				charB2 = charB1 - '0';
			// Convert the digit to decimal and add it to the
			// decimalValue of n
			decimalValue = decimalValue * b1 + charB2;
		}
		
		// Converting the decimal value to base b2:
		// A number is converted from decimal to another base
		// by continuously dividing by the base and recording 
		// the remainder until the quotient is zero. The number in the
		// new base is the remainders, with the last remainder
		// being the left-most digit.
		
		// While the quotient is NOT zero:
		while (decimalValue != 0) {
			// If the remainder is a digit < 10, simply add it to
			// the left side of the new number.
			if (decimalValue % b2 < 10) 
				output = Integer.toString(decimalValue % b2) + output;
			// If the remainder is >= 10, add a character with the
			// corresponding value to the new number. (A = 10, B = 11, C = 12, ...)
			else
				output = (char)((decimalValue % b2)+55) + output;
			// Divide by the new base again
			decimalValue /= b2;
		}
		return output;
	}
}