README_EN.md

318. Maximum Product of Word Lengths

中文文档

Description

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.

Solutions

Python3

class Solution:
    def maxProduct(self, words: List[str]) -> int:
        n = len(words)
        masks = [0] * n
        for i, word in enumerate(words):
            for c in word:
                masks[i] |= (1 << (ord(c) - ord('a')))
        ans = 0
        for i in range(n - 1):
            for j in range(i + 1, n):
                if (masks[i] & masks[j]) == 0:
                    ans = max(ans, len(words[i]) * len(words[j]))
        return ans

Java

class Solution {
    public int maxProduct(String[] words) {
        int n = words.length;
        int[] masks = new int[n];
        for (int i = 0; i < n; ++i) {
            for (char c : words[i].toCharArray()) {
                masks[i] |= (1 << (c - 'a'));
            }
        }
        int ans = 0;
        for (int i = 0; i < n - 1; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if ((masks[i] & masks[j]) == 0) {
                    ans = Math.max(ans, words[i].length() * words[j].length());
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int n = words.size();
        vector<int> masks(n);
        for (int i = 0; i < n; ++i)
            for (char c : words[i])
                masks[i] |= (1 << (c - 'a'));
        int ans = 0;
        for (int i = 0; i < n - 1; ++i)
            for (int j = i + 1; j < n; ++j)
                if ((masks[i] & masks[j]) == 0)
                    ans = max(ans, (int) (words[i].size() * words[j].size()));
        return ans;
    }
};

Go

func maxProduct(words []string) int {
	n := len(words)
	masks := make([]int, n)
	for i, word := range words {
		for _, c := range word {
			masks[i] |= (1 << (c - 'a'))
		}
	}
	ans := 0
	for i := 0; i < n-1; i++ {
		for j := i + 1; j < n; j++ {
			if (masks[i] & masks[j]) == 0 {
				ans = max(ans, len(words[i])*len(words[j]))
			}
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

...