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