Reaching from one string to another using given dictionary - arrays

Question

For this question, a dictionary was given and two strings also given, it was basically asked to reach from one string to another one just using the words in dictionary, and only one letter can be changed at a time. I came up with this solution. There were some corner cases that my code can not handle. Can you help to find all the corner cases to make this code prettier?
public static int findNumberOfSteps(String start, String end , HashSet<String> dict){
if( start == null || end == null || dict.isEmpty()){
throw new IllegalArgumentException();
}
dict.add(end);
Queue<String> wordHolder = new LinkedList<>();
Queue<Integer> distanceCount = new LinkedList<Integer>();
wordHolder.add(start);
distanceCount.add(1);
int result = Integer.MAX_VALUE;
while (!wordHolder.isEmpty()){
String currentWord = wordHolder.poll();
int currDistance = distanceCount.poll();
if(currentWord.equals(end)){
int result = currDistance;
return result;
}
for (int i = 0 ; i < currentWord.length() ; i++){
char[] charCurrentWord = currentWord.toCharArray();
for ( char c = 'a' ; c <= 'z' ; c++){
charCurrentWord[i] = c;
String newWord = new String(charCurrentWord);
if (dict.contains(newWord)){
wordHolder.add(newWord);
distanceCount.add(currDistance+1);
dict.remove(newWord);
}
}
}
}
return 0;
}

Answer 1

There are a couple of problems in the code. The first problem is in this code
if(currentWord.equals(end)){
result = Math.min(result, currDistance);
}
Note that when you reach the end word, that code updates the result, but then the code is going to search for ways to change the end word to something else. That's a huge waste of time, the code should continue with the while(!wordHolder.isEmpy()) loop after the end is found.
The second problem is in this code
if (dict.contains(newWord)){
wordHolder.add(newWord);
distanceCount.add(currDistance+1);
dict.remove(newWord);
}
Note that if newWord is equal to the end word, then that code removes the end word from the dictionary, which means that you'll never find the end word again.
The solution to both problems is to check for the end word inside that if statement. When the end is found, don't add it to the wordHolder and don't remove it from the dictionary.
if (dict.contains(newWord)){
if(newWord.equals(end)){
result = Math.min(result, currDistance+1);
}
else{
wordHolder.add(newWord);
distanceCount.add(currDistance+1);
dict.remove(newWord);
}
}