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LeetCode

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Problem

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.

Code

class Solution {
    private int encode(String s) {
        int res = 0;
        for (char c : s.toCharArray()) {
            res |= (1 << (c - 'a'));
        }

        return res;
    }

    public int maxProduct(String[] words) {
        int[] encodes = new int[words.length];

        for (int i = 0; i < words.length; i++){
            encodes[i] = encode(words[i]);
        }

        int res = 0;

        for(int i = 0; i < encodes.length; i++) {
            for(int j = 0; j < encodes.length; j++){
                if((encodes[i] & encodes[j]) == 0) {
                    res = Math.max(res, words[i].length() * words[j].length());
                }
            }
        }

        return res;
    }
}

class Solution:
    def maxProduct(self, words: List[str]) -> int:
        encodes = [0] * len(words)

        for i, word in enumerate(words):
            enc = 0
            for c in word:
                enc |= 1 << (ord(c) - ord('a'))

            encodes[i] = enc

        res = 0

        for i, word1 in enumerate(words):
            for j, word2 in enumerate(words):
                if encodes[i] & encodes[j] == 0:
                    res = max(res, len(word1) * len(word2))

        return res