318. 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:

Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16 
Explanation: The two words can be "abcw", "xtfn".

Example 2:

Input: ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4 
Explanation: The two words can be "ab", "cd".

Example 3:

Input: ["a","aa","aaa","aaaa"]
Output: 0 
Explanation: No such pair of words.

class Solution {
    public int maxProduct(String[] words) {
        int l = words.length;
        int res = 0;
        if(l == 0) return res;
        // if(l == 1) return 
        for(int i = 0; i < l - 1; i++){
            for(int j = i + 1; j <l; j++){
                String t1 = words[i];
                String t2 = words[j];
                boolean t = true;
                for(int m = 0; m < t1.length(); m++){
                    for(int n = 0; n < t2.length(); n++){
                        if(t1.charAt(m) == t2.charAt(n)){
                            t = false;
                            break;
                        }
                    }
                    if(t == false) break;
                    
                }
                if(t == true) res = Math.max(res, t1.length() * t2.length());
            }
        }
        return res;
    }
}

brute force

public class Solution {
    public int maxProduct(String[] words) {
        final int n = words.length;
        final boolean[][] hashset = new boolean[n][26];

        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < words[i].length(); ++j) {
                hashset[i][words[i].charAt(j) - 'a'] = true;
            }
        }

        int result = 0;
        for (int i = 0; i < n-1; ++i) {
            for (int j = i + 1; j < n; ++j) {
                boolean hasCommon = false;
                for (int k = 0; k < 26; ++k) {
                    if (hashset[i][k] && hashset[j][k]) {
                        hasCommon = true;
                        break;
                    }
                }
                int tmp = words[i].length() * words[j].length();
                if (!hasCommon && tmp > result) {
                    result = tmp;
                }
            }
        }
        return result;
    }
    // private static final int ALPHABET_SIZE = 26;
}

 

class Solution {
    public int maxProduct(String[] words) {
        int n = words.length;
        int[] val = new int[n];
        
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < words[i].length(); j++) {               
                val[i] |= (1 << words[i].charAt(j) - 'a'); //val[i] means the "value" of current word by calculating "OR" operation through each character, which denoted by the index of each char.
            }
        }
        
        int res = 0;
        for(int i = 0; i < n; i++) {
            for(int j = i + 1; j < n; j++) {
                if((val[i] & val[j]) == 0 && res < words[i].length() * words[j].length()) {
                    res = words[i].length() * words[j].length();
                }
            }
        }
        return res;
    }
}

用bit做，用一个array来存这个string的value，用二进制表示。然后比较两个string的二进制value，没有共同的就更新res