318

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.

Companies:
Google

Related Topics:
Bit Manipulation

Solution 1.

// OJ: https://leetcode.com/problems/maximum-product-of-word-lengths/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int maxProduct(vector<string>& A) {
        int N = A.size();
        if (N == 0) return 0;
        vector<int> bit(N);
        for (int i = 0; i < N; ++i) {
            for (char c : A[i]) {
                bit[i] |= (1 << (c - 'a'));
            }
        }
        int ans = 0;
        for (int i = 0; i < N; ++i) {
            for (int j = i + 1; j < N; ++j) {
                if ((bit[i] & bit[j]) == 0) {
                    ans = max(ans, (int)(A[i].size() * A[j].size()));
                }
            }
        }
        return ans;
    }
};