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:

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.

Solution:

The idea to compare whether two words has the same letter is to use bit representation for each word, or int. 

a is the last bit of the int and z is the 26th bit in the int.

CODE:

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

Another similar question is that there is a matrix contains 0s and 1s, find the number of rectangle where all the four corner is 1.

if we check to rows, 

111011,

011110

and do the & operation.

011010, there are 3 1s in the result, so the number would be pick 2 out of 3 is 6;