I'm trying to find all the 3, 4, 5, and 6 letter words given 6 letters. I am finding them by comparing every combination of the 6 letters to an ArrayListArrayList of words called sortedDictionarysortedDictionary. I have worked on the code a good bit to get it to this point. I tested how many six letter words are checked and got 720 which is good because 65432*1=720 which means I am not checking any words twice. I can't make it faster by getting rid of duplicate checks because I have already gotten rid of them all.
Note that sortedDictionarysortedDictionary only contains about 27 hundred words.
for(int l1 = 0; l1 < 6; l1++){
for(int l2 = 0; l2 < 6; l2++){
if(l2 != l1)
for(int l3 = 0; l3 < 6; l3++){
if(l3 != l1 && l3 != l2){
if(sortedDictionary.contains(""+anagramCharacters[l1]+anagramCharacters[l2]+anagramCharacters[l3]))
anagram_words.add(""+anagramCharacters[l1]+anagramCharacters[l2]+anagramCharacters[l3]);
for(int l4 = 0; l4 < 6; l4++){
if(l4 != l1 && l4 != l2 && l4 != l3){
if(sortedDictionary.contains(""+anagramCharacters[l1]+anagramCharacters[l2]+anagramCharacters[l3]+anagramCharacters[l4]))
anagram_words.add(""+anagramCharacters[l1]+anagramCharacters[l2]+anagramCharacters[l3]+anagramCharacters[l4]);
for(int l5 = 0; l5 < 6; l5++){
if(l5 != l1 && l5 != l2 && l5 != l3 && l5 != l4){
if(sortedDictionary.contains(""+anagramCharacters[l1]+anagramCharacters[l2]+anagramCharacters[l3]+anagramCharacters[l4]+anagramCharacters[l5]))
anagram_words.add(""+anagramCharacters[l1]+anagramCharacters[l2]+anagramCharacters[l3]+anagramCharacters[l4]+anagramCharacters[l5]);
for(int l6 = 0; l6 < 6; l6++){
if(l6 != l1 && l6 != l2 && l6 != l3 && l6 != l4 && l6 != l5)
if(sortedDictionary.contains(""+anagramCharacters[l1]+anagramCharacters[l2]+anagramCharacters[l3]+anagramCharacters[l4]+anagramCharacters[l5]+anagramCharacters[l6]))
anagram_words.add(""+anagramCharacters[l1]+anagramCharacters[l2]+anagramCharacters[l3]+anagramCharacters[l4]+anagramCharacters[l5]+anagramCharacters[l6]);
}
}
}
}
}
}
}
}
}
My solution, still probably not the best written code but it reduce the loading time from about 1-2 seconds to nearly instant (no noticeable wait time,time; didn't actually test how long it was).
for(int i = 0; i < sortedDictionary.size(); i++){
for(int index = 0; index < anagram.length(); index++)
anagramCharacters[index] = anagram.charAt(index);
forloop:
for(int i2 = 0; i2 < sortedDictionary.get(i).length(); i2++){
for(int i3 = 0; i3 < anagramCharacters.length; i3++){
if(sortedDictionary.get(i).charAt(i2) == anagramCharacters[i3]){
anagramCharacters[i3] = 0;
break;
}
else if(i3 == anagramCharacters.length-1)
break forloop;
}
if(i2 == sortedDictionary.get(i).length()-1)
anagram_words.add(sortedDictionary.get(i));
}
}
