LeetCode Q318: Maximum Product of Word Lengths

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:

Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be "abcw", "xtfn".

Example 2:

Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be "ab", "cd".

Example 3:

Given ["a", "aa", "aaa", "aaaa"]
Return 0
No such pair of words.

Solution:

Use bit manipulation to speed up. When  char string appears in a question and is asked to do something related to check whether identical string, most likely you need to use bit manipulation.

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

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