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:

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.

分析：

怎么判断两个string没有共同的letter。 我们可以用int的每一个位来表示是否某一个letter出现在某一个字符串中。如果两个字符串的int值AND以后为0，很明显，它们是没有公共letter.

public class Solution {
    public int maxProduct(String[] words) {
        if (words == null || words.length <= 1) return 0;
        
        int[] wordInfo = new int[words.length];
        for (int i = 0; i < words.length; i++) {
            for (int j = 0; j < words[i].length(); j++) {
                // each letter is the 1 bit in the number 
                wordInfo[i] = wordInfo[i] | (1 << words[i].charAt(j) - 'a');
            }
        }
        int maxProduct = 0;
        for (int i = 0; i < words.length; i++) {
            for (int j = i + 1; j < words.length; j++) {
                if (((wordInfo[i] & wordInfo[j]) == 0)) {
                    maxProduct = Math.max(maxProduct, words[i].length() * words[j].length());
                }
            }
        }

        return maxProduct;
    }
}