318. Maximum Product of Word Lengths

Posted 2020-08-01 蜃利的阴影下

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.

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 Bit Manipulation

 

public class Solution {
    public int maxProduct(String[] words) {
        int max = 0;
        //sort the array based on string length in a descending order.
        Arrays.sort(words, new Comparator<String>(){
           @Override
           public int compare(String a, String b)
           {
               return b.length() - a.length();
           }
        });
        
        //get the spectrum for each word.
        int[] spectrum = new int[words.length];
        
        for(int w = 0; w<words.length; ++w)
        {
            String word = words[w];
            for(int i = 0; i<word.length(); ++i)
            {
                spectrum[w] |= 1<<word.charAt(i) - ‘a‘;  //1<<32 <==> 1<<0. 32 is the cycle.
            }
        }
        
        for(int i = 0; i<words.length-1; ++i)
        {
            int l = words[i].length();
            if(l*l <= max)
                break;
            for(int j = i+1; j<words.length; ++j)
            {
                if((spectrum[i] & spectrum[j]) == 0)
                {
                    max = Math.max(max, l*words[j].length());
                    break;
                }
            }
        }
        return max;
    }
}