# Python solution with detailed explanation

• Solution

Maximum Product of Word Lengths https://leetcode.com/problems/maximum-product-of-word-lengths/

Algorithm

• Simple brute force solution is to test if every pair for similarity and take the product is they are not similar. Return the maximum product if several dissimilar pairs are found otherwise return 0.
• First optimization is to run the two for loops from** i = (0 to N-1) and j = (i+1, N-1)**. This is lesser than N^2.

How do we find if two words are similar?

• First approach will be to take two words, put all characters into a set and test for membership in that set for the other words. Can we do better? Can we pre-process somehow?
• Second approach will be to prepprocess each word and generate a unique sign for it. Take a boolean array of 26. Mark the position for every character in this word as True. Now traverse this array in order and generate a unique signature. O(26 + len(word)). Can we still do better?
• Can we use bit-wise manipulation? int32 is 32 bits. There are 26 letters. Set a bit for every character. How do you test if two words have no similar letters? Just AND them. Testing them now becomes a constant time operation
``````class Solution(object):
def sign(self, word):
value = 0
for c in word:
value = value | (1 << (ord(c)-97))
return value

def maxProduct(self, words):
"""
:type words: List[str]
:rtype: int
"""
signature = [self.sign(x) for x in words]
max_product, N = 0, len(words)
for i in range(N):
for j in range(i+1, N):
if signature[i] & signature[j] == 0:
max_product = max(max_product, len(words[i])*len(words[j]))
return max_product
``````

Use set as a signature
Yes another way is to use sets in an interesting manner.

``````class Solution(object):
def maxProduct(self, words):
"""
:type words: List[str]
:rtype: int
"""
signature = {w:set(w) for w in words}
max_product, N = 0, len(words)
for i in range(N):
for j in range(i+1, N):
# Intersection of two sets
if bool(signature[words[i]] & signature[words[j]]) == False:
max_product = max(max_product, len(words[i])*len(words[j]))
return max_product
``````

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