Finding longest substring in alphabetical order

Question

EDIT: I am aware that a question with similar task was already asked in SO but I\'m interested to find out the problem in this specific piece of code. I am

Answer 1

Simple and easy to understand:

s = "abcbcd"    #The original string

l = len(s)    #The length of the original string

maxlenstr = s[0]    #maximum length sub-string, taking the first letter of original string as value.

curlenstr = s[0]    #current length sub-string, taking the first letter of original string as value.

for i in range(1,l):    #in range, the l is not counted. 

    if s[i] >= s[i-1]:    #If current letter is greater or equal to previous letter,
        curlenstr += s[i] #add the current letter to current length sub-string
    else:        
        curlenstr = s[i]  #otherwise, take the current letter as current length sub-string

    if len(curlenstr) > len(maxlenstr): #if current cub-string's length is greater than max one,
            maxlenstr = curlenstr;      #take current one as max one.

print("Longest substring in alphabetical order is:", maxlenstr)

Answer 2

I agree with @Abhijit about the power of itertools.groupby() but I took a simpler approach to (ab)using it and avoided the boundary case problems:

from itertools import groupby

LENGTH, LETTERS = 0, 1

def longest_sorted(string):
    longest_length, longest_letters = 0, []
    key, previous_letter = 0, chr(0)

    def keyfunc(letter):
        nonlocal key, previous_letter
        if letter < previous_letter:
            key += 1

        previous_letter = letter
        return key

    for _, group in groupby(string, keyfunc):
        letters = list(group)
        length = len(letters)

        if length > longest_length:
            longest_length, longest_letters = length, letters

    return ''.join(longest_letters)

print(longest_sorted('hixwluvyhzzzdgd'))
print(longest_sorted('eseoojlsuai'))
print(longest_sorted('drurotsxjehlwfwgygygxz'))
print(longest_sorted('abcdefghijklmnopqrstuvwxyz'))

OUTPUT

> python3 test.py
luvy
jlsu
ehlw
abcdefghijklmnopqrstuvwxyz
>

Answer 3

Here. This does what you want. One pass, no need for recursion.

def find_longest_substring_in_alphabetical_order(s):
    groups = []
    cur_longest = ''
    prev_char = ''
    for c in s.lower():
        if prev_char and c < prev_char:
            groups.append(cur_longest)
            cur_longest = c
        else:
            cur_longest += c
        prev_char = c
    return max(groups, key=len) if groups else s

Using it:

>>> find_longest_substring_in_alphabetical_order('hixwluvyhzzzdgd')
'luvy'
>>> find_longest_substring_in_alphabetical_order('eseoojlsuai')
'jlsu'
>>> find_longest_substring_in_alphabetical_order('drurotsxjehlwfwgygygxz')
'ehlw'

Note: It will probably break on strange characters, has only been tested with the inputs you suggested. Since this is a "homework" question, I will leave you with the solution as is, though there is still some optimization to be done, I wanted to leave it a little bit understandable.

Answer 4

Assuming this is from Edx course: till this question, we haven't taught anything about strings and their advanced operations in python So, I would simply go through the looping and conditional statements

string =""             #taking a plain string to represent the then generated string
present =""             #the present/current longest string
for i in range(len(s)): #not len(s)-1 because that totally skips last value
    j = i+1            
    if j>= len(s):    
        j=i           #using s[i+1] simply throws an error of not having index
    if s[i] <= s[j]:  #comparing the now and next value
        string += s[i] #concatinating string if above condition is satisied
    elif len(string) != 0 and s[i] > s[j]: #don't want to lose the last value
        string += s[i] #now since s[i] > s[j] #last one will be printed
        if len(string) > len(present): #1 > 0 so from there we get to store many values
            present = string #swapping to largest string
        string = ""
    if len(string) > len(present): #to swap from if statement
        present = string
if present == s[len(s)-1]: #if no alphabet is in order then first one is to be the output
    present = s[0]
print('Longest substring in alphabetical order is:' + present)

Answer 5

s = 'gkuencgybsbezzilbfg'
x = s.lower()
y = ''
z = [] #creating an empty listing which will get filled

for i in range(0,len(x)):
    if i == len(x)-1:
       y = y + str(x[i])
       z.append(y)
       break 
    a = x[i] <= x[i+1]
    if a == True:
        y = y + str(x[i])
    else: 
        y = y + str(x[i])
        z.append(y)  # fill the list 
        y = ''
# search of 1st longest string        
L = len(max(z,key=len))      # key=len takes length in consideration 
for i in range(0,len(z)):
    a = len(z[i])
    if a == L:   
        print 'Longest substring in alphabetical order is:' + str(z[i])
        break