leetcode maximumproductofwordlengths

import java.util.HashMap;

public class  {
  
   * Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters.
   * You may assume that each word will contain only lower case letters.
   * If no such two words exist, return 0.
   *
   * Example 1:
   *
   * Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
   * Output: 16
   * Explanation: The two words can be "abcw", "xtfn".
   * Example 2:
   *
   * Input: ["a","ab","abc","d","cd","bcd","abcd"]
   * Output: 4
   * Explanation: The two words can be "ab", "cd".
   * Example 3:
   *
   * Input: ["a","aa","aaa","aaaa"]
   * Output: 0
   * Explanation: No such pair of words.
   */
  public int maxProduct(String[] words) {
    int[] masks = getMask(words);
    int max = 0;
    for (int i=1; i<words.length; i++){
      for (int j=0; j<i; j++){
        if ((masks[i] & masks[j]) == 0){
          max = Math.max(max, words[i].length()+words[j].length());
        }
      }
    }
    return max;
  }

  public HashMap<Character, Integer> getBinaryMapping(){
    HashMap<Character, Integer> mapping = new HashMap<>();
    for (int i=0; i<26; i++){
      mapping.put((char) ('a'+i), (1 << i));
    }
    return mapping;
  }

  public int[] getMask(String[] words){
    int[] masks = new int[words.length];
    HashMap<Character, Integer> mapping = getBinaryMapping();
    for (int i=0; i<words.length; i++){
      int mask = 0;
      for (char c: words[i].toCharArray()){
        mask |= mapping.get(c);
      }
      masks[i] = mask;
    }
    return masks;
  }
}