318_Maximum_Product_of_Word_Lengths

Leetcode 318 : Maximum Product of Word Lengths
Detailed video explanation : https://youtu.be/E8Ctj36CzuY
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C++:
----
class Solution {
public:
    int maxProduct(vector<string>& words) {
        int n = words.size();
        vector<int> word_int(n, 0);
        vector<int> word_len(n, 0);
        for(int i = 0; i < n; ++i){
            string w = words[i];
            int _len = w.length();
            int w_int = 0;
            for(int j = 0; j < _len; ++j)
                w_int |= 1 << (w[j] - 'a');
            word_int[i] = w_int;
            word_len[i] = _len;
        }
        
        int max_product = 0;
        for(int i = 0; i < n-1; ++i){
            for(int j = i+1; j < n; ++j){
                if((word_int[i] & word_int[j]) == 0)
                    max_product = max(max_product, word_len[i]*word_len[j]);
            }
        }
        return max_product;
    }
};




Java:
-----
class Solution {
    public int maxProduct(String[] words) {
        int n = words.length;
        int[] word_int = new int[n];
        int[] word_len = new int[n];
        for(int i = 0; i < n; ++i){
            String w = words[i];
            int _len = w.length();
            int w_int = 0;
            for(int j = 0; j < _len; ++j)
                w_int |= 1 << (w.charAt(j) - 'a');
            word_int[i] = w_int;
            word_len[i] = _len;
        }
        
        int max_product = 0;
        for(int i = 0; i < n-1; ++i){
            for(int j = i+1; j < n; ++j){
                if((word_int[i] & word_int[j]) == 0)
                    max_product = Math.max(max_product, word_len[i]*word_len[j]);
            }
        }
        return max_product;
    }
}



Python3:
-------
class Solution:
    def maxProduct(self, words: List[str]) -> int:
        n = len(words)
        word_int, word_len = [0]*n, [0]*n
        for i in range(n):
            w = words[i]
            _len = len(w)
            w_int = 0
            for j in range(_len):
                w_int |= 1 << (ord(w[j]) - ord('a'))
            word_int[i] = w_int
            word_len[i] = _len
        
        max_product = 0
        for i in range(n-1):
            for j in range(i+1, n):
                if((word_int[i] & word_int[j]) == 0):
                    max_product = max(max_product, word_len[i]*word_len[j])
        return max_product