Mystery sum with placeholder digits

I came across this coding problem.


Back in primary school, maybe you were sometimes asked to solve a
fill-in-the-blank sum - or "mystery sum" - in which certain digits are
removed and you had to figure out what they originally were.


Now, it's time for serious business: sums with no digits, where every
digit is replaced by a letter. Within each seperate sum, a letter
replaces one digit and it is always the same letter for that one
digit.


For example, in the sum below: A+BB=ACC The aim is to figure out which
digits are hidden behind the letters A, B and C so that the sum turns
out correct. Here, the solution is that A replaces 1, B replaces 9 and
C replaces 0; resulting in: 1+99=100


Finally, there is the constraint that the length of the given word is
equal to the length of the hidden number. Thus, in the sum
CHEVAL+VACHE=OISEAU, CHEVAL is indeed a 6 digit number: the C (more
precisely the first letter of any word) cannot correspond to a zero.


For each given sum, there is only one possible solution to the
problem. You are asked to write the result value of the sum to the
standard output.


INPUT: One line : a sum in the form WORD1+WORD2=RESULT, in which each
word is a sequence of capital letters.


OUTPUT: The result of the solved sum.


CONSTRAINTS: Each sum has 2 to 20 words, each made up of 1 to 20
capital letters

I solved it using brute force checking against each permutation of assigned integers to the unique characters. Can anybody suggest a better strategy or algorithm?

```
public static void mysterySum(String equation) {
String[] breakUp = equation.split("=");
String[] lhs = breakUp[0].split("\\+");
String rhs = breakUp[1];

HashSet charSet = new HashSet();
for (char charVal : equation.toCharArray()) {
if (!(charVal == '+' || charVal == '=')) {
charSet.add(new Character(charVal));
}
}

char[] refCharVals = new char[10];

int i = 0;
for (Character charVal : charSet) {
re

As for a review of the code I would suggest to avoid acronyms (also recommend, as @200_success did, to better indent the code, in case this is how your code is indented).

As for the solution, I would strongly avoid brute force ones. This is a cryptarithmetic problem and, if you work only on base 10, you have 10! possible assignments, but if you generalize (for any base) then the problem is NP-complete. I would suggest to look for solutions for cryptarithmetic problems (usually they are described on texts/sources on AI).

StackExchange Code Review Q#92546, answer score: 2