Nsryjdtyk

I have a "solution" to a problem but I am not satisfied with it. I will first pose the question, show my solution, and then explain my issues with my solution.

Question:
In a computer network, data must be transmitted between different nodes (devices). Unfortunately, this is not an error-proof
process. Transmission problems can garble a message, leading to errors in the data that is received. Luckily, the errors are
(usually) randomly distributed, so they will occur in different places each time a message is transmitted. By transmitting
the same message multiple times and comparing the results, we can gather data that will help us to reconstruct the original
message. To do this, we examine each position of each message copy. The value that occurs most frequently in a specific
position is (usually) the correct value for that position.
For example, consider the following repeated/retransmitted message (in this example, spaces have been inserted between
each letter for clarity):

f e l n o w o w l d

h e j w o a p r o d

g e l q h w k b p y

h p r s e w o r l d

h e l l b m o r l d

h e g l o x l r h d

h h l l o w e r k u

h t l k q w x s l d

In the first column/position, we have 1 f, 1 g, and 6 h’s. Since "h" is the most common value for the first letter, we can guess
that the first letter of the original message was most likely "h". The second position has 5 e’s, 1 h, 1 p, and 1 t, so "e" is the
most likely second character. Repeat this process for each position in the message copies. In this case, the original message
turns out to be "helloworld".
Complete the recoverMessage() function, which takes a list of equal-length strings as its single argument. You may
assume that each string ONLY contains lowercase letters; there are no spaces, digits, uppercase letters, or punctuation
symbols in any of the source strings The function returns a single string whose length is equal to that of any one of the
source strings.

My solution:

def recoverMessage(x):

alphabet = {"a" : 0, "b" : 0, "c":0,"d":0,"e":0,"f":0,"g":0,"h":0,"i":0,

            "j":0,"k":0,"l":0,"m":0,"n":0,"o":0,"p":0,"q":0,"r":0,"s":0,"t":0,

            "u":0,"v":0,"w":0,"x":0,"y":0,"z":0}

unscrambled = ""

sidestring = ""

for i in range(len(x[0])):

    for j in range(len(x)):

        alphabet[x[j][i]] = alphabet[x[j][i]] + 1

    y = max(alphabet.values())

    for letters in alphabet:

        if alphabet.get(letters) == y:

            sidestring = sidestring + letters

    unscrambled = unscrambled + sidestring[0]



    sidestring = ""

    alphabet = {"a" : 0, "b" : 0, "c":0,"d":0,"e":0,"f":0,"g":0,"h":0,"i":0,

            "j":0,"k":0,"l":0,"m":0,"n":0,"o":0,"p":0,"q":0,"r":0,"s":0,"t":0,

            "u":0,"v":0,"w":0,"x":0,"y":0,"z":0}



return (unscrambled)

The issue I have with my solution is that it doesn't output all possible solutions. For example, if you manually do this for the function call

recoverMessage(["citputejncienje", "cebautersmjynzt",

"comcntewsciunwb", "foxuuterscienfe",
"jomvuzerociiavn", "tojeotesstfdiqe",
"computezsciryce"]),

you should get computerscienje OR
computerscience. My issue is in my code I incorporate the max function. But a tie could result. How can I save all these letters that come in a tie and output all possible "correct" words.