find number of places an alphabet to be shifted to make it alaphabetical order in python

Question

Here let us consider a string 'images'

original string is str = 'images' alphabetical order is str = 'aegims'

to get this alphabetical order we shifted 'a' two places forward (i.e a) and 'e' to three places forward(i.e ae) and 'g' to one place forward (i.e aeg) and 'i' to three places backward(i.e aegi)

so when user enters a letter we need to display the number of places it shifted

for suppose if user enters letter 'e' we need to print output as 3 places shifted

Answer 1

If I understood your question correctly, you want to code something like this:

class Word():
    def __init__(self, word='') -> None:
        self.initial = word
        self.final = ''
        self.char_shift = {}
        self.__alphabetize()
        self.__calculate_shift()
    
    def __alphabetize(self) -> None:
        word = list(self.initial)
        #sort 
        counter = 0
        sorted = False
        
        while not sorted:
            #assume already sorted
            sorted = True
            for i in range(len(word)-1-counter):
                #check if actually sorted
                if word[i] > word[i+1]:
                    #set sorted
                    sorted = False
                    #swop
                    temp = word[i]
                    word[i] = word[i+1]
                    word[i+1] = temp
                
            #increase loop count
            counter += 1
        
        for char in word: self.final += char      
    
    def __calculate_shift(self) -> None:  
        for char in self.initial:
            self.char_shift[char] = abs(self.final.index(char) - self.initial.index(char))

It made more sense to make it a class, though functional would also work. Here are my inputs and outputs:

word1 = Word('images')
print(word1.final)
print(word1.initial)
print(word1.char_shift['e'])

Output

aegims
images
3