Leetcode Maximum Product of Word Lengths problem solution

In this Leetcode Maximum Product of Word Lengths problem solution, You are given a string array of 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.

Problem solution in Python.

class Solution(object):

    def maxProduct(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        word_bit = [0 for _ in range(len(words))]
        maxLen = 0
        for i, word in enumerate(words):
            for c in word:
                mask = 1 << ord(c) - ord('a')
                word_bit[i] |= mask

        for i in range(0, len(words) - 1):
            for j in range(i + 1, len(words)):
                if word_bit[i] & word_bit[j] == 0:
                    maxLen = max(len(words[i]) * len(words[j]), maxLen)

        return maxLen

Problem solution in Java.

public int maxProduct(String[] words) {
    int n = words.length, max = 0;
    int[] mask = new int[n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < words[i].length(); j++)
            mask[i] |= 1 << (words[i].charAt(j) - 'a');
        for (int j = 0; j < i; j++)
            if ((mask[j] & mask[i]) == 0)
                max = Math.max(max, words[i].length() * words[j].length());
    }
    return max;
}

Problem solution in C++.

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int res=INT_MIN;
        
        for(int i=0;i<words.size();i++){
           
            for(int j=0;j<words.size() and j!=i;j++){
                if(words[i].find_first_of(words[j])==string::npos){
                    int l=words[i].length()*words[j].length();
                    res=max(res,l);
                }
            }
        }
        return max(res,0);
    }
};

Problem solution in C.

#include <stdlib.h>
#include <string.h>

struct wordData
{
    unsigned int length;
    unsigned int bitMask;
};

int maxProduct(char ** words, int wordsSize){
    int idx, idx2, maxProd = 0;
    struct wordData *pWordData;
    
    pWordData = malloc(sizeof(struct wordData) * wordsSize);
    for (idx = 0; idx < wordsSize; idx++)
    {
        char *pWord;  
        
        pWord = words[idx];
        pWordData[idx].length = 0;
        pWordData[idx].bitMask = 0;
        while ('' != *pWord)
        {
            pWordData[idx].length++;
            pWordData[idx].bitMask |= (1 << (*pWord - 'a'));
            pWord++;
        }
    }
    for (idx = 0; idx < wordsSize; idx++)
    {
        for (idx2 = idx + 1; idx2 < wordsSize; idx2++)
        {
            if (0 == (pWordData[idx].bitMask & pWordData[idx2].bitMask))
            {
                int prod;
                
                prod = pWordData[idx].length * pWordData[idx2].length;
                maxProd = (prod > maxProd) ? prod : maxProd;
            }
        }
    }
    
    free(pWordData);
    
    return maxProd;
}

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