[LeetCode]318. Maximum Product of Word Lengths

Medium

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:
Given [“abcw”, “baz”, “foo”, “bar”, “xtfn”, “abcdef”]
Return 16
The two words can be “abcw”, “xtfn”.

Example 2:
Given [“a”, “ab”, “abc”, “d”, “cd”, “bcd”, “abcd”]
Return 4
The two words can be “ab”, “cd”.

Example 3:
Given [“a”, “aa”, “aaa”, “aaaa”]
Return 0
No such pair of words.

35ms:

     public int maxProduct(String[] words) {        int max = 0;        int[] bytes = new int[words.length];        for(int i=0;i<words.length;i++){            int val = 0;            for(int j=0;j<words[i].length();j++){                val |= 1<<(words[i].charAt(j)-'a');            }            bytes[i] = val;        }        for(int i=0; i<bytes.length; i++){            for(int j=i+1; j<bytes.length; j++){                if((bytes[i] & bytes[j])==0)max = Math.max(max,words[i].length()*words[j].length());            }        }        return max;      }