[LC] 318. Maximum Product of Word Lengths

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.

class Solution {
    public int maxProduct(String[] words) {
        int[] checker = new int[words.length];
        if (words == null || words.length == 0) {
            return 0;
        }
        int res = 0;
        
        for (int i = 0; i < words.length; i++) {
            String word = words[i];
            for (int j = 0; j < word.length(); j++) {
                char curChar = word.charAt(j);
                checker[i] |= 1 << curChar - 'a';
            }
        }
        
        for (int i = 0; i < words.length - 1; i++) {
            for (int j = i + 1; j < words.length; j++) {
                if ((checker[i] & checker[j]) == 0) {
                    res = Math.max(res, words[i].length() * words[j].length());
                }
            }
        }
        return res;
    }
}

public class Solution {
  public int largestProduct(String[] dict) {
    // Write your solution here
    Arrays.sort(dict, new Comparator<String>(){
      @Override
      public int compare(String a, String b) {
        return b.length() - a.length();
      }
    });
    int[] arr = new int[dict.length];
    for (int i = 0; i < dict.length; i++) {
      for (int j = 0; j < dict[i].length(); j++) {
        arr[i] |= 1 << dict[i].charAt(j) - 'a';
      }
    }
    int res = 0;
    for (int i = 1; i < dict.length; i++) {
      for (int j = 0; j < i; j++) {
        if (dict[i].length() * dict[j].length() <= res) {
          break;
        }
        if ((arr[i] & arr[j]) == 0) {
          res = dict[i].length() * dict[j].length();
        }
      }
    }
    return res;
  }
}