Introduction

In this blog post, we will explore how to solve the LeetCode problem "318" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a string array 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. Example 1: Input: words = ["abcw","baz","foo","bar","xtfn","abcdef"] Output: 16 Explanation: The two words can be "abcw", "xtfn". Example 2: Input: words = ["a","ab","abc","d","cd","bcd","abcd"] Output: 4 Explanation: The two words can be "ab", "cd". Example 3: Input: words = ["a","aa","aaa","aaaa"] Output: 0 Explanation: No such pair of words. Constraints: 2 <= words.length <= 1000 1 <= words[i].length <= 1000 words[i] consists only of lowercase English letters.

Explanation

Here's the breakdown of the optimal solution:

• Bit Manipulation: Represent each word as a bitmask, where each bit corresponds to a letter of the alphabet. This allows for efficient checking of common letters using bitwise AND.
• Precomputation: Calculate the bitmask for each word beforehand to avoid redundant calculations within the nested loops.

• Optimized Iteration: Iterate through all pairs of words, checking for common letters using the bitmasks. If no common letters are found, update the maximum product.

• Runtime Complexity: O(N² + NL), where N is the number of words and L is the total number of characters across all words (for creating bitmasks). Storage Complexity: O(N), where N is the number of words (for storing bitmasks).

Code

    def max_product(words):
    """
    Finds the maximum product of lengths of two words that do not share common letters.

    Args:
      words: A list of strings.

    Returns:
      The maximum product of lengths, or 0 if no such pair exists.
    """

    n = len(words)
    masks = [0] * n

    # Create bitmasks for each word
    for i in range(n):
        for char in words[i]:
            masks[i] |= 1 << (ord(char) - ord('a'))

    max_prod = 0
    for i in range(n):
        for j in range(i + 1, n):
            if (masks[i] & masks[j]) == 0:
                max_prod = max(max_prod, len(words[i]) * len(words[j]))

    return max_prod