LeetCode [318] Maximum Product of Word Lengths

 318. Maximum Product of Word Lengths

Medium

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:

Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16 
Explanation: The two words can be "abcw", "xtfn".

Example 2:

Input: ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4 
Explanation: The two words can be "ab", "cd".

Example 3:

Input: ["a","aa","aaa","aaaa"]
Output: 0 
Explanation: No such pair of words.

 

Constraints:

• 0 <= words.length <= 10^3
• 0 <= words[i].length <= 10^3
• words[i] consists only of lowercase English letters.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

class Solution { public static int maxProduct(String[] words) { if (words == null || words.length == 0) return 0; int len = words.length; int[] value = new int[len]; for (int i = 0; i < len; i++) { String tmp = words[i]; value[i] = 0; for (int j = 0; j < tmp.length(); j++) { value[i] |= 1 << (tmp.charAt(j) - 'a'); } } int maxProduct = 0; for (int i = 0; i < len; i++) for (int j = i + 1; j < len; j++) { if ((value[i] & value[j]) == 0 && (words[i].length() * words[j].length() > maxProduct)) maxProduct = words[i].length() * words[j].length(); } return maxProduct; } }