LeetCode-in-Java

318. Maximum Product of Word Lengths

Medium

Given a string array words, return the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. If no such two words exist, return 0.

Example 1:

Input: words = [“abcw”,”baz”,”foo”,”bar”,”xtfn”,”abcdef”]

Output: 16

Explanation: The two words can be “abcw”, “xtfn”.

Example 2:

Input: words = [“a”,”ab”,”abc”,”d”,”cd”,”bcd”,”abcd”]

Output: 4

Explanation: The two words can be “ab”, “cd”.

Example 3:

Input: words = [“a”,”aa”,”aaa”,”aaaa”]

Output: 0

Explanation: No such pair of words.

Constraints:

• 2 <= words.length <= 1000
• 1 <= words[i].length <= 1000
• words[i] consists only of lowercase English letters.

Solution

public class Solution {
    public int maxProduct(String[] words) {
        int n = words.length;
        int res = 0;
        int[] masks = new int[n];
        for (int i = 0; i < n; i++) {
            masks[i] = getMask(words[i]);
        }
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if ((masks[i] & masks[j]) == 0) {
                    res = Math.max(res, words[i].length() * words[j].length());
                }
            }
        }
        return res;
    }

    private int getMask(String s) {
        int mask = 0;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            mask = mask | (1 << (c - 'a'));
        }
        return mask;
    }
}

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