refactor 318

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_318.java

@@ -1,27 +1,5 @@
 package com.fishercoder.solutions;
 
-import java.util.Arrays;
-import java.util.HashSet;
-import java.util.Set;
-
-/**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.*/
 public class _318 {
     public static class Solution1 {
         //Inspired by this awesome post: https://discuss.leetcode.com/topic/35539/java-easy-version-to-understand
@@ -38,15 +16,15 @@ public int maxProduct(String[] words) {
                 String word = words[i];
                 for (int j = 0; j < words[i].length(); j++) {
                     values[i] |= 1 << (word.charAt(j)
-                        - 'a');//the reason for left shift by this number "word.charAt(j) -'a'" is for 'a', otherwise 'a' - 'a' will be zero and 'a' will be missed out.
+                            - 'a');//the reason for left shift by this number "word.charAt(j) -'a'" is for 'a', otherwise 'a' - 'a' will be zero and 'a' will be missed out.
                 }
             }
             int maxProduct = 0;
             for (int i = 0; i < words.length; i++) {
                 for (int j = 0; j < words.length; j++) {
                     //check if values[i] AND values[j] equals to zero, this means they share NO common chars
                     if ((values[i] & values[j]) == 0
-                        && words[i].length() * words[j].length() > maxProduct) {
+                            && words[i].length() * words[j].length() > maxProduct) {
                         maxProduct = words[i].length() * words[j].length();
                     }
                 }