Solution is still tricky, one has to take a decission what way to lean in my eyes. Because of the 3/7 ratio, do we have to see a 3/8 as a too long space between the characters or a too short space between words.
In this solution 1110111 will result in a 'M', 11100111/111000111/1110000111/11100000111/11100000111/11100000000111 (2,3,4,5,6,8 0's) into an 'I' and 1110000000111 (7 0's) into a 'T', 111000000000111 and any longer seuquences of 0's (9.. 0's) into an 'EE'

If the sequence is not optimal it will fail, just put an extra 0 in the HEY JUDE to see the result. String replacement might be clever but does not catch the variations of the keyer. Which is the problem with all computer-based CW-decoders