Find the longest substring in alphabetical order

Question

I have this code that I found on another topic, but it sorts the substring by contiguous characters and not by alphabetical order. How do I correct it for alphabetical order? It prints out lk, and I want to print ccl. Thanks

ps: I'm a beginner in python

s = 'cyqfjhcclkbxpbojgkar'
from itertools import count

def long_alphabet(input_string):
    maxsubstr = input_string[0:0] # empty slice (to accept subclasses of str)
    for start in range(len(input_string)): # O(n)
        for end in count(start + len(maxsubstr) + 1): # O(m)
            substr = input_string[start:end] # O(m)
            if len(set(substr)) != (end - start): # found duplicates or EOS
                break
            if (ord(max(sorted(substr))) - ord(min(sorted(substr))) + 1) == len(substr):
                maxsubstr = substr
    return maxsubstr

bla = (long_alphabet(s))
print "Longest substring in alphabetical order is: %s" %bla

Answer 1

You can improve your algorithm by noticing that the string can be broken into runs of ordered substrings of maximal length. Any ordered substring must be contained in one of these runs

This allows you to just iterate once through the string O(n)

def longest_substring(string):
    curr, subs = '', ''
    for char in string:
        if not curr or char >= curr[-1]:
            curr += char
        else:
            curr, subs = '', max(curr, subs, key=len)
    return max(curr, subs, key=len)

Answer 2

s = 'cyqfjhcclkbxpbojgkar'
longest = ""
max =""

for i in range(len(s) -1):
    if(s[i] <= s[i+1] ):
       longest = longest + s[i]
       if(i==len(s) -2):
           longest = longest + s[i+1]
    else:
        longest  = longest + s[i]        
        if(len(longest) > len(max)):
            max = longest
        longest = ""        

if(len(s) == 1):
    longest = s


if(len(longest) > len(max)):
    print("Longest substring in alphabetical order is: " + longest)
else:
    print("Longest substring in alphabetical order is: " + max)