# In-Depth Explanation and Solution in Python using Bit Manipulation

• ``````class Solution(object):
def maxProduct(self, words):
maxProduct = 0
binaryWords = []
for word in words:
binaryWords.append(self.wordToBinary(word))
for i in xrange(len(words)):
for j in xrange(i+1,len(words)):
#If words do not contain like characters
if(not (binaryWords[i] & binaryWords[j])):
maxProduct = max(maxProduct, len(words[i]) * len(words[j]))
return maxProduct

def wordToBinary(self,word):
num = 0
for char in word:
num |= (1 << ord(char)-ord('a'))
return num
``````

The only trick to this problem is finding out how to efficiently determine whether or not two words share any common letters. In order to do this, we create a new array to contain binary representation of the words. For example, the right-most bit would act as a switch to represent whether or not the word contains an 'a', the bit to the left of that a 'b', and so forth. Once our strings are represented as numbers, we can apply bitwise and (&) to any two words to find out if they share any letters in common. If the result of the & statement is 0, the two binary representations of our words do not share any common 1s. Below is an example:

word: "ab"
binary representation: 11 or 0011

word: "cd"
binary representation: 1100

0011 & 1100 = 0, therefor these two words do not share any common letters.

Once you've figured that out, the rest is just a matter of looping through all the words and multiplying the lengths of the ones that satisfy the above condition, while keeping track of the max.

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