# Python Solution --- Easy to understand!

• ``````def maxProduct(self, words):
n, res = len(words), 0
word_bits = [0 for i in range(n)]
# generate a 26 bit mask for every word, each set to 1 if a character is set
for i in range(n):
for c in words[i]:
mask = 1 << (ord(c) - 96)
# for each word, compre to every other word
for i in range(n):
for j in range(i+1, n):
# if the AND of the two bits are zero, then no overlapping characters
if word_bits[i] & word_bits[j] == 0:
res = max(res, len(words[i]) * len(words[j]))
return res
``````

Keep is to use a bit vector to keep track of which characters are set for each word. For example, if we have word `ade`, then we can represent this with `10011`, where a is the first bit, and e is the last bit etc. There are two zero bits to represent the missing `b, c`.

Now to check if two words have the same characters, we just need to do an & of their vectors. If the result is 0, then there is no overlapping characters.

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