Leetcode: 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.

想到String的char 查重，最好的方法就是用bit manipulation

注意O(N^2)的iteration里面可以适当pruning

public class Solution {
    public int maxProduct(String[] words) {
        if (words==null || words.length==0) return 0;
        int max = 0;
        int[] bitmasks = new int[words.length]; 
        Arrays.sort(words, new Comparator<String>(){
            public int compare(String s1, String s2) {
                return s2.length()-s1.length();
            }
        });
        
        for (int i=0; i<words.length; i++) {
            for (int j=0; j<words[i].length(); j++) {
                bitmasks[i] |= 1<<((int)(words[i].charAt(j) - 'a'));
            }
        }
        
        for (int i=0; i<words.length-1; i++) {
            for (int j=i+1; j<words.length; j++) {
                if (words[i].length() * words[j].length() <= max) break; //pruning
                if ((bitmasks[i] & bitmasks[j]) != 0) continue;
                max = words[i].length() * words[j].length();
                break;//pruning
            }
        }
        return max;
    }
}