Maximum Product Of Word Lengths

                                            
class Solution {
public:
    int maxProduct(vector<string>& words) {
        int n = words.size();
        vector<int> masks(n);
        vector<int> lens(n);

        for (int i = 0; i < n; i++) {
            int mask = 0;
            for (char c : words[i]) {
                mask |= 1 << (c - 'a');
            }
            masks[i] = mask;
            lens[i] = words[i].length();
        }

        int maxProd = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if ((masks[i] & masks[j]) == 0) {
                    maxProd = max(maxProd, lens[i] * lens[j]);
                }
            }
        }

        return maxProd;
    }
};

                                            
class Solution {
    public int maxProduct(String[] words) {
        int max = 0;
        
        int[] bitMasks = new int[words.length];
        for (int i = 0; i < words.length; i++) {
            for (char c : words[i].toCharArray()) {
                bitMasks[i] |= 1 << (c - 'a');
            }
        }
        
        for (int i = 0; i < words.length; i++) {
            for (int j = i + 1; j < words.length; j++) {
                if ((bitMasks[i] & bitMasks[j]) == 0) {
                    max = Math.max(max, words[i].length() * words[j].length());
                }
            }
        }
        
        return max;
    }
}

                                            
class Solution(object):
    def maxProduct(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        def has_common_letters(w1, w2):
            return any(letter in w2 for letter in w1)

        max_prod = 0
        for i in range(len(words)):
            for j in range(i+1, len(words)):
                if not has_common_letters(words[i], words[j]):
                    max_prod = max(max_prod, len(words[i]) * len(words[j]))
        
        return max_prod

                                            
int maxProduct(char** words, int wordsSize) {
    int max = 0;
    
    for (int i = 0; i < wordsSize; i++) {
        for (int j = i + 1; j < wordsSize; j++) {
            int flag = 0;
            for (int k = 0; words[i][k] != '\0'; k++) {
                for (int l = 0; words[j][l] != '\0'; l++) {
                    if (words[i][k] == words[j][l]) {
                        flag = 1;
                        break;
                    }
                }
                if (flag == 1) {
                    break;
                }
            }
            
            if (flag == 0) {
                int product = strlen(words[i]) * strlen(words[j]);
                if (product > max) {
                    max = product;
                }
            }
        }
    }
    
    return max;
}

To solve this coding challenge, we need to compute the maximum product of lengths of any two words in a given array of words such that the two words do not share any common letters. Let's break down the solution step-by-step:

# Explanation

Identify the Problem Constraints

We're given an array of words.
Each word is composed of lowercase English letters.
We need to find two words without common letters and compute their length product.
Return the maximum such product. If no such pairs are found, return 0.

Approach to Solve the Problem

For each pair of words, we need to check if they share any common letters.
If there's no common letter, compute the product of their lengths.
Track the maximum product found during these computations.
Efficiently checking common letters and storing the maximum product are key.

Optimizations to Consider

Directly checking each pair could be expensive; hence bit manipulation or other set-based optimizations are usually more efficient.
For simplicity here, basic checks and nested loops will be described, but improvements could be suggested later.

Step-by-Step Explanation/Detailed Steps in Pseudocode:

Iterate over every possible pair of words.
For each pair, check if they have common letters:

Convert each word to a set of characters.
Use set intersection to check for common letters.

If no common letters are found, compute the product of the lengths of the two words.
Keep track of the maximum product found.
Return the maximum product after all pairs are checked.

Pseudocode with Comments:

                                            
# Function to determine if two words share any common letters
function hasCommonLetters(word1, word2):
    # Convert each word to a set of characters
    set1 = set of characters from word1
    set2 = set of characters from word2
    
    # Check intersection of sets; if non-empty, there is a common letter
    if there are common elements in set1 and set2:
        return true
    else:
        return false

# Main function to find maximum product of lengths of two non-overlapping words
function maxProduct(wordsList):
    # Initialize variable to store the maximum product
    maxProductValue = 0
    
    # Outer loop to go through each word in the list
    for i from 0 to length(wordsList) - 1:
        # Inner loop to go through each subsequent word in the list
        for j from i + 1 to length(wordsList):
            # If words do not share common letters
            if not hasCommonLetters(wordsList[i], wordsList[j]):
                # Calculate the product of their lengths
                product = length(wordsList[i]) * length(wordsList[j])
                
                # Update maximum product if the current product is larger
                if product > maxProductValue:
                    maxProductValue = product
    
    # Return the largest product found
    return maxProductValue

# Example usage:
wordsInput = ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
result = maxProduct(wordsInput)
print(result)  # Expected Output: 16, due to "abcw" and "xtfn"

In the provided pseudocode:

The
hasCommonLetters

function checks if two words share any letters using set intersections.
The
maxProduct

function iterates over all pairs of words, calculates their product if they don’t share letters, and keeps track of the maximum product found.
This basic approach provides clarity and correctness, but remember that further optimizations are possible, like using bitmasks.

By following the detailed explanation and the corresponding pseudocode, one can understand the methodology and logic behind solving the given coding challenge.