How do you solve Maximum Product of Word Lengths on LeetCode?

Maximum Product of Word Lengths (LeetCode #318) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Using bit manipulation can significantly speed up the process of checking for common letters.

What is the brute force solution for Maximum Product of Word Lengths?

The brute force approach for Maximum Product of Word Lengths is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of words to see if they share any common letters. If they don't, we calculate the product of their lengths and keep track of the maximum product found. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Maximum Product of Word Lengths?

Maximum Product of Word Lengths uses the following concepts: Array, String, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Two Pointers.

What are the interview tips for Maximum Product of Word Lengths?

Always clarify the problem and constraints before jumping into coding. Think about the time complexity of your solution and whether it can be improved. Practice explaining your thought process as you code, as this is often part of the interview.

What are common mistakes when solving Maximum Product of Word Lengths?

Not considering edge cases where words have common letters. Overlooking the efficiency of using bit manipulation for character comparison.

#318

Maximum Product of Word Lengths

Medium

ArrayStringBit ManipulationBit ManipulationTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

The optimal solution uses bit manipulation to represent each word as a bitmask. This allows us to quickly check for common letters using bitwise operations, significantly reducing the time complexity.

⚙️

Algorithm

4 steps

1Step 1: Create an array to store the bitmask for each word, where each bit represents a letter in the alphabet.
2Step 2: For each word, compute its bitmask and store it in the array.
3Step 3: Use two nested loops to compare the bitmasks of each pair of words. If the bitwise AND of the two masks is 0, they share no common letters.
4Step 4: Calculate the product of their lengths and update the maximum product accordingly.

solution.py20 lines

1# Full working Python code
2words = ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
3
4def maxProduct(words):
5    n = len(words)
6    masks = [0] * n
7    lengths = [0] * n
8    for i in range(n):
9        for char in words[i]:
10            masks[i] |= (1 << (ord(char) - ord('a')))
11        lengths[i] = len(words[i])
12
13    max_product = 0
14    for i in range(n):
15        for j in range(i + 1, n):
16            if masks[i] & masks[j] == 0:
17                max_product = max(max_product, lengths[i] * lengths[j])
18    return max_product
19
20print(maxProduct(words))

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but the bit manipulation allows us to check for common letters in constant time. The space complexity is O(n) for storing the masks and lengths.

1Using bit manipulation can significantly speed up the process of checking for common letters.
2Understanding how to represent characters as bits can lead to more efficient solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.