318. Maximum Product of Word Lengths

Problem:

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.

Analysis:

Convert string to bit, just record the occurrence of char. abc should be 00111, ab should be 00011. If there is a char in string, the char - 'a' th bit in mask integer should be 1. 

If two strings have exact same letters, use &, the result will be 1, otherwise 0.

Solution:

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