# Maximum Product Of Word Lengths

Maximum Product Of Word Lengths bit manipulation

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.
``````

Maximum Product Of Word Lengths Solution

``````public class MaximumProductOfWordLengths {
public int maxProduct(String[] words) {
if(words.length == 0 || words == null) {
return 0;
}

int length = words.length;
int[] value = new int[length];
int max = 0;

for(int i = 0; i < length; i++) {
String temp = words[i];

value[i] = 0;

for(int j = 0; j < temp.length(); j++) {
value[i] |= 1 << (temp.charAt(j) - 'a');
}
}

for(int i = 0; i < length; i++) {
for(int j = 1; j < length; j++) {
if((value[i] & value[j]) == 0 && (words[i].length() * words[j].length()) > max) {
max = words[i].length() * words[j].length();
}
}
}

return max;
}
}``````