Maximum Product of Word Lengths – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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.

C++ Maximum Product of Word Lengths LeetCode Solution

int maxProduct(vector<string>& words) {
    unordered_map<int,int> maxlen;
    int result = 0;
    for (string word : words) {
        int mask = 0;
        for (char c : word)
            mask |= 1 << (c - 'a');
        maxlen[mask] = max(maxlen[mask], (int) word.size());
        for (auto maskAndLen : maxlen)
            if (!(mask & maskAndLen.first))
                result = max(result, (int) word.size() * maskAndLen.second);
    }
    return result;
}

Java Maximum Product of Word Lengths LeetCode 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;
}
 

Python 3 Maximum Product of Word Lengths LeetCode Solution

class Solution(object):
    def maxProduct(self, words):
        d = {}
        for w in words:
            mask = 0
            for c in set(w):
                mask |= (1 << (ord(c) - 97))
            d[mask] = max(d.get(mask, 0), len(w))
        return max([d[x] * d[y] for x in d for y in d if not x & y] or [0])

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