Leetcode #318: Maximum Product of Word Lengths
In this guide, we solve Leetcode #318 Maximum Product of Word Lengths in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Bit Manipulation, Array, String
Intuition
The problem structure lets us track state with bitwise operations.
Bit operations are constant time and avoid extra memory.
Approach
Apply XOR/AND/OR and shifts to maintain the required invariant.
Aggregate the result in a single pass.
Steps:
- Identify a bitwise invariant.
- Combine values with bit operations.
- Return the aggregated result.
Example
Input: words = ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16
Explanation: The two words can be "abcw", "xtfn".
Python Solution
class Solution:
def maxProduct(self, words: List[str]) -> int:
mask = [0] * len(words)
ans = 0
for i, s in enumerate(words):
for c in s:
mask[i] |= 1 << (ord(c) - ord("a"))
for j, t in enumerate(words[:i]):
if (mask[i] & mask[j]) == 0:
ans = max(ans, len(s) * len(t))
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.