Develop Reference

How to take a string of digits and find out if the digits are increasing?

For this function I am supposed to take in 2 parameters that are both strings. The first parameter is any positive digit and the second is a digit that is supposed to split the other digit into groupings. This is an example of the code:
def split_tester(N, d):
# Your code should include dosctrings and the body of the function
if N.isdigit() and d.isdigit():
N1 = int(N)
d1 = int(d)
substring = ""
new_sub = ""
retsub = ""
for i in N:
substring = substring + i
if len(substring) == d1:
new_sub = new_sub +substring+ " "
substring = ""
print(new_sub)
So in the shell it would print 12 34 when split_tester("1234","2"). My question is how can I show that the 34 is greater than the 12 when these are cast as strings and are in the same variable?

Simplifying your function a little bit:
def split_tester(N, d):
d_num= int(d)
num_list = [int(N[i:i+d_num]) for i in range(0, len(N), d_num)]
for i in num_list:
print(i, end=" ")
split_tester("1234", "2")
You can perform your desired comparison (if 12 is greater than 34 etc.) on the numbers of the output list: num_list

Brute Force password generator for md5? [duplicate]

I would like to make a alphabetical list for an application similar to an excel worksheet.
A user would input number of cells and I would like to generate list.
For example a user needs 54 cells. Then I would generate
'a','b','c',...,'z','aa','ab','ac',...,'az', 'ba','bb'
I can generate the list from [ref]
from string import ascii_lowercase
L = list(ascii_lowercase)
How do i stitch it together?
A similar question for PHP has been asked here. Does some one have the python equivalent?

Use itertools.product.
from string import ascii_lowercase
import itertools
def iter_all_strings():
for size in itertools.count(1):
for s in itertools.product(ascii_lowercase, repeat=size):
yield "".join(s)
for s in iter_all_strings():
print(s)
if s == 'bb':
break
Result:
a
b
c
d
e
...
y
z
aa
ab
ac
...
ay
az
ba
bb
This has the added benefit of going well beyond two-letter combinations. If you need a million strings, it will happily give you three and four and five letter strings.
Bonus style tip: if you don't like having an explicit break inside the bottom loop, you can use islice to make the loop terminate on its own:
for s in itertools.islice(iter_all_strings(), 54):
print s

You can use a list comprehension.
from string import ascii_lowercase
L = list(ascii_lowercase) + [letter1+letter2 for letter1 in ascii_lowercase for letter2 in ascii_lowercase]

Following #Kevin 's answer :
from string import ascii_lowercase
import itertools
# define the generator itself
def iter_all_strings():
size = 1
while True:
for s in itertools.product(ascii_lowercase, repeat=size):
yield "".join(s)
size +=1
The code below enables one to generate strings, that can be used to generate unique labels for example.
# define the generator handler
gen = iter_all_strings()
def label_gen():
for s in gen:
return s
# call it whenever needed
print label_gen()
print label_gen()
print label_gen()

I've ended up doing my own.
I think it can create any number of letters.
def AA(n, s):
r = n % 26
r = r if r > 0 else 26
n = (n - r) / 26
s = chr(64 + r) + s
if n > 26:
s = AA(n, s)
elif n > 0:
s = chr(64 + n) + s
return s
n = quantity | r = remaining (26 letters A-Z) | s = string
To print the list :
def uprint(nc):
for x in range(1, nc + 1):
print AA(x,'').lower()
Used VBA before convert to python :
Function AA(n, s)
r = n Mod 26
r = IIf(r > 0, r, 26)
n = (n - r) / 26
s = Chr(64 + r) & s
If n > 26 Then
s = AA(n, s)
ElseIf n > 0 Then
s = Chr(64 + n) & s
End If
AA = s
End Function

Using neo's insight on a while loop.
For a given iterable with chars in ascending order. 'abcd...'.
n is the Nth position of the representation starting with 1 as the first position.
def char_label(n, chars):
indexes = []
while n:
residual = n % len(chars)
if residual == 0:
residual = len(chars)
indexes.append(residual)
n = (n - residual)
n = n // len(chars)
indexes.reverse()
label = ''
for i in indexes:
label += chars[i-1]
return label
Later you can print a list of the range n of the 'labels' you need using a for loop:
my_chrs = 'abc'
n = 15
for i in range(1, n+1):
print(char_label(i, my_chrs))
or build a list comprehension etc...

Print the set of xl cell range of lowercase and uppercase charterers
Upper_case:
from string import ascii_uppercase
import itertools
def iter_range_strings(start_colu):
for size in itertools.count(1):
for string in itertools.product(ascii_uppercase, repeat=size):
yield "".join(string)
input_colume_range = ['A', 'B']
input_row_range= [1,2]
for row in iter_range_strings(input_colume_range[0]):
for colum in range(int(input_row_range[0]), int(input_row_range[1]+1)):
print(str(row)+ str(colum))
if row == input_colume_range[1]:
break
Result:
A1
A2
B1
B2

In two lines (plus an import):
from string import ascii_uppercase as ABC
count = 100
ABC+=' '
[(ABC[x[0]] + ABC[x[1]]).strip() for i in range(count) if (x:= divmod(i-26, 26))]
Wrap it in a function/lambda if you need to reuse.

code:
alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
for i in range(len(alphabet)):
for a in range(len(alphabet)):
print(alphabet[i] + alphabet[a])
result:
aa
ab
ac
ad
ae
af
ag
ah
ai
aj
ak
al
am
...

Add a start index to a string index generator

I'm currently learning to create generators and to use itertools. So I decided to make a string index generator, but I'd like to add some parameters such as a "start index" allowing to define where to start generating the indexes.
I came up with this ugly solution which can be very long and not efficient with large indexes:
import itertools
import string
class StringIndex(object):
'''
Generator that create string indexes in form:
A, B, C ... Z, AA, AB, AC ... ZZ, AAA, AAB, etc.
Arguments:
- startIndex = string; default = ''; start increment for the generator.
- mode = 'lower' or 'upper'; default = 'upper'; is the output index in
lower or upper case.
'''
def __init__(self, startIndex = '', mode = 'upper'):
if mode == 'lower':
self.letters = string.ascii_lowercase
elif mode == 'upper':
self.letters = string.ascii_uppercase
else:
cmds.error ('Wrong output mode, expected "lower" or "upper", ' +
'got {}'.format(mode))
if startIndex != '':
if not all(i in self.letters for i in startIndex):
cmds.error ('Illegal characters in start index; allowed ' +
'characters are: {}'.format(self.letters))
self.startIndex = startIndex
def getIndex(self):
'''
Returns:
- string; current string index
'''
startIndexOk = False
x = 1
while True:
strIdMaker = itertools.product(self.letters, repeat = x)
for stringList in strIdMaker:
index = ''.join([s for s in stringList])
# Here is the part to simpify
if self.startIndex:
if index == self.startIndex:
startIndexOk = True
if not startIndexOk:
continue
###
yield index
x += 1
Any advice or improvement is welcome. Thank you!
EDIT:
The start index must be a string!

You would have to do the arithmetic (in base 26) yourself to avoid looping over itertools.product. But you can at least set x=len(self.startIndex) or 1!

Old (incorrect) answer
If you would do it without itertools (assuming you start with a single letter), you could do the following:
letters = 'abcdefghijklmnopqrstuvwxyz'
def getIndex(start, case):
lets = list(letters.lower()) if case == 'lower' else list(letters.upper())
# default is 'upper', but can also be an elif
for r in xrange(0,10):
for l in lets[start:]:
if l.lower() == 'z':
start = 0
yield ''.join(lets[:r])+l
I run until max 10 rows of letters are created, but you could ofcourse use an infinite while loop such that it can be called forever.
Correct answer
I found the solution in a different way: I used a base 26 number translator (based on (and fixxed since it didn't work perfectly): http://quora.com/How-do-I-write-a-program-in-Python-that-can-convert-an-integer-from-one-base-to-another)
I uses itertools.count() to count and just loops over all the possibilities.
The code:
import time
from itertools import count
def toAlph(x, letters):
div = 26
r = '' if x > 0 else letters[0]
while x > 0:
r = letters[x % div] + r
if (x // div == 1) and (x % div == 0):
r = letters[0] + r
break
else:
x //= div
return r
def getIndex(start, case='upper'):
alphabet = 'abcdefghijklmnopqrstuvwxyz'
letters = alphabet.upper() if case == 'upper' else alphabet
started = False
for num in count(0,1):
l = toAlph(num, letters)
if l == start:
started = True
if started:
yield l
iterator = getIndex('AA')
for i in iterator:
print(i)
time.sleep(0.1)

How to make a continuous alphabetic list python (from a-z then from aa, ab, ac etc)

I would like to make a alphabetical list for an application similar to an excel worksheet.
A user would input number of cells and I would like to generate list.
For example a user needs 54 cells. Then I would generate
'a','b','c',...,'z','aa','ab','ac',...,'az', 'ba','bb'
I can generate the list from [ref]
from string import ascii_lowercase
L = list(ascii_lowercase)
How do i stitch it together?
A similar question for PHP has been asked here. Does some one have the python equivalent?

Use itertools.product.
from string import ascii_lowercase
import itertools
def iter_all_strings():
for size in itertools.count(1):
for s in itertools.product(ascii_lowercase, repeat=size):
yield "".join(s)
for s in iter_all_strings():
print(s)
if s == 'bb':
break
Result:
a
b
c
d
e
...
y
z
aa
ab
ac
...
ay
az
ba
bb
This has the added benefit of going well beyond two-letter combinations. If you need a million strings, it will happily give you three and four and five letter strings.
Bonus style tip: if you don't like having an explicit break inside the bottom loop, you can use islice to make the loop terminate on its own:
for s in itertools.islice(iter_all_strings(), 54):
print s

You can use a list comprehension.
from string import ascii_lowercase
L = list(ascii_lowercase) + [letter1+letter2 for letter1 in ascii_lowercase for letter2 in ascii_lowercase]

Following #Kevin 's answer :
from string import ascii_lowercase
import itertools
# define the generator itself
def iter_all_strings():
size = 1
while True:
for s in itertools.product(ascii_lowercase, repeat=size):
yield "".join(s)
size +=1
The code below enables one to generate strings, that can be used to generate unique labels for example.
# define the generator handler
gen = iter_all_strings()
def label_gen():
for s in gen:
return s
# call it whenever needed
print label_gen()
print label_gen()
print label_gen()

I've ended up doing my own.
I think it can create any number of letters.
def AA(n, s):
r = n % 26
r = r if r > 0 else 26
n = (n - r) / 26
s = chr(64 + r) + s
if n > 26:
s = AA(n, s)
elif n > 0:
s = chr(64 + n) + s
return s
n = quantity | r = remaining (26 letters A-Z) | s = string
To print the list :
def uprint(nc):
for x in range(1, nc + 1):
print AA(x,'').lower()
Used VBA before convert to python :
Function AA(n, s)
r = n Mod 26
r = IIf(r > 0, r, 26)
n = (n - r) / 26
s = Chr(64 + r) & s
If n > 26 Then
s = AA(n, s)
ElseIf n > 0 Then
s = Chr(64 + n) & s
End If
AA = s
End Function

Using neo's insight on a while loop.
For a given iterable with chars in ascending order. 'abcd...'.
n is the Nth position of the representation starting with 1 as the first position.
def char_label(n, chars):
indexes = []
while n:
residual = n % len(chars)
if residual == 0:
residual = len(chars)
indexes.append(residual)
n = (n - residual)
n = n // len(chars)
indexes.reverse()
label = ''
for i in indexes:
label += chars[i-1]
return label
Later you can print a list of the range n of the 'labels' you need using a for loop:
my_chrs = 'abc'
n = 15
for i in range(1, n+1):
print(char_label(i, my_chrs))
or build a list comprehension etc...

Print the set of xl cell range of lowercase and uppercase charterers
Upper_case:
from string import ascii_uppercase
import itertools
def iter_range_strings(start_colu):
for size in itertools.count(1):
for string in itertools.product(ascii_uppercase, repeat=size):
yield "".join(string)
input_colume_range = ['A', 'B']
input_row_range= [1,2]
for row in iter_range_strings(input_colume_range[0]):
for colum in range(int(input_row_range[0]), int(input_row_range[1]+1)):
print(str(row)+ str(colum))
if row == input_colume_range[1]:
break
Result:
A1
A2
B1
B2

In two lines (plus an import):
from string import ascii_uppercase as ABC
count = 100
ABC+=' '
[(ABC[x[0]] + ABC[x[1]]).strip() for i in range(count) if (x:= divmod(i-26, 26))]
Wrap it in a function/lambda if you need to reuse.

code:
alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
for i in range(len(alphabet)):
for a in range(len(alphabet)):
print(alphabet[i] + alphabet[a])
result:
aa
ab
ac
ad
ae
af
ag
ah
ai
aj
ak
al
am
...

Finding longest substring in alphabetical order

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 also aware that this problem can be solved without using recursion.
The task is to write a program which will find (and print) the longest sub-string in which the letters occur in alphabetical order. If more than 1 equally long sequences were found, then the first one should be printed. For example, the output for a string abczabcd will be abcz.
I have solved this problem with recursion which seemed to pass my manual tests. However when I run an automated tests set which generate random strings, I have noticed that in some cases, the output is incorrect. For example:
if s = 'hixwluvyhzzzdgd', the output is hix instead of luvy
if s = 'eseoojlsuai', the output is eoo instead of jlsu
if s = 'drurotsxjehlwfwgygygxz', the output is dru instead of ehlw
After some time struggling, I couldn't figure out what is so special about these strings that causes the bug.
This is my code:
pos = 0
maxLen = 0
startPos = 0
endPos = 0
def last_pos(pos):
if pos < (len(s) - 1):
if s[pos + 1] >= s[pos]:
pos += 1
if pos == len(s)-1:
return len(s)
else:
return last_pos(pos)
return pos
for i in range(len(s)):
if last_pos(i+1) != None:
diff = last_pos(i) - i
if diff - 1 > maxLen:
maxLen = diff
startPos = i
endPos = startPos + diff
print s[startPos:endPos+1]

There are many things to improve in your code but making minimum changes so as to make it work. The problem is you should have if last_pos(i) != None: in your for loop (i instead of i+1) and you should compare diff (not diff - 1) against maxLen. Please read other answers to learn how to do it better.
for i in range(len(s)):
if last_pos(i) != None:
diff = last_pos(i) - i + 1
if diff > maxLen:
maxLen = diff
startPos = i
endPos = startPos + diff - 1

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.

You can use nested for loops, slicing and sorted. If the string is not all lower-case then you can convert the sub-strings to lower-case before comparing using str.lower:
def solve(strs):
maxx = ''
for i in xrange(len(strs)):
for j in xrange(i+1, len(strs)):
s = strs[i:j+1]
if ''.join(sorted(s)) == s:
maxx = max(maxx, s, key=len)
else:
break
return maxx
Output:
>>> solve('hixwluvyhzzzdgd')
'luvy'
>>> solve('eseoojlsuai')
'jlsu'
>>> solve('drurotsxjehlwfwgygygxz')
'ehlw'

Python has a powerful builtin package itertools and a wonderful function within groupby
An intuitive use of the Key function can give immense mileage.
In this particular case, you just have to keep a track of order change and group the sequence accordingly. The only exception is the boundary case which you have to handle separately
Code
def find_long_cons_sub(s):
class Key(object):
'''
The Key function returns
1: For Increasing Sequence
0: For Decreasing Sequence
'''
def __init__(self):
self.last_char = None
def __call__(self, char):
resp = True
if self.last_char:
resp = self.last_char < char
self.last_char = char
return resp
def find_substring(groups):
'''
The Boundary Case is when an increasing sequence
starts just after the Decresing Sequence. This causes
the first character to be in the previous group.
If you do not want to handle the Boundary Case
seperately, you have to mak the Key function a bit
complicated to flag the start of increasing sequence'''
yield next(groups)
try:
while True:
yield next(groups)[-1:] + next(groups)
except StopIteration:
pass
groups = (list(g) for k, g in groupby(s, key = Key()) if k)
#Just determine the maximum sequence based on length
return ''.join(max(find_substring(groups), key = len))
Result
>>> find_long_cons_sub('drurotsxjehlwfwgygygxz')
'ehlw'
>>> find_long_cons_sub('eseoojlsuai')
'jlsu'
>>> find_long_cons_sub('hixwluvyhzzzdgd')
'luvy'

Simple and easy.
Code :
s = 'hixwluvyhzzzdgd'
r,p,t = '','',''
for c in s:
if p <= c:
t += c
p = c
else:
if len(t) > len(r):
r = t
t,p = c,c
if len(t) > len(r):
r = t
print 'Longest substring in alphabetical order is: ' + r
Output :
Longest substring in alphabetical order which appeared first: luvy

Here is a single pass solution with a fast loop. It reads each character only once. Inside the loop operations are limited to
1 string comparison (1 char x 1 char)
1 integer increment
2 integer subtractions
1 integer comparison
1 to 3 integer assignments
1 string assignment
No containers are used. No function calls are made. The empty string is handled without special-case code. All character codes, including chr(0), are properly handled. If there is a tie for the longest alphabetical substring, the function returns the first winning substring it encountered. Case is ignored for purposes of alphabetization, but case is preserved in the output substring.
def longest_alphabetical_substring(string):
start, end = 0, 0 # range of current alphabetical string
START, END = 0, 0 # range of longest alphabetical string yet found
prev = chr(0) # previous character
for char in string.lower(): # scan string ignoring case
if char < prev: # is character out of alphabetical order?
start = end # if so, start a new substring
end += 1 # either way, increment substring length
if end - start > END - START: # found new longest?
START, END = start, end # if so, update longest
prev = char # remember previous character
return string[START : END] # return longest alphabetical substring
Result
>>> longest_alphabetical_substring('drurotsxjehlwfwgygygxz')
'ehlw'
>>> longest_alphabetical_substring('eseoojlsuai')
'jlsu'
>>> longest_alphabetical_substring('hixwluvyhzzzdgd')
'luvy'
>>>

a lot more looping, but it gets the job done
s = raw_input("Enter string")
fin=""
s_pos =0
while s_pos < len(s):
n=1
lng=" "
for c in s[s_pos:]:
if c >= lng[n-1]:
lng+=c
n+=1
else :
break
if len(lng) > len(fin):
fin= lng`enter code here`
s_pos+=1
print "Longest string: " + fin

def find_longest_order():
`enter code here`arr = []
`enter code here`now_long = ''
prev_char = ''
for char in s.lower():
if prev_char and char < prev_char:
arr.append(now_long)
now_long = char
else:
now_long += char
prev_char = char
if len(now_long) == len(s):
return now_long
else:
return max(arr, key=len)
def main():
print 'Longest substring in alphabetical order is: ' + find_longest_order()
main()

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)

s = input("insert some string: ")
start = 0
end = 0
temp = ""
while end+1 <len(s):
while end+1 <len(s) and s[end+1] >= s[end]:
end += 1
if len(s[start:end+1]) > len(temp):
temp = s[start:end+1]
end +=1
start = end
print("longest ordered part is: "+temp)

I suppose this is problem set question for CS6.00.1x on EDX. Here is what I came up with.
s = raw_input("Enter the string: ")
longest_sub = ""
last_longest = ""
for i in range(len(s)):
if len(last_longest) > 0:
if last_longest[-1] <= s[i]:
last_longest += s[i]
else:
last_longest = s[i]
else:
last_longest = s[i]
if len(last_longest) > len(longest_sub):
longest_sub = last_longest
print(longest_sub)

I came up with this solution
def longest_sorted_string(s):
max_string = ''
for i in range(len(s)):
for j in range(i+1, len(s)+1):
string = s[i:j]
arr = list(string)
if sorted(string) == arr and len(max_string) < len(string):
max_string = string
return max_string

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)

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
>

s = 'azcbobobegghakl'
i=1
subs=s[0]
subs2=s[0]
while(i<len(s)):
j=i
while(j<len(s)):
if(s[j]>=s[j-1]):
subs+=s[j]
j+=1
else:
subs=subs.replace(subs[:len(subs)],s[i])
break
if(len(subs)>len(subs2)):
subs2=subs2.replace(subs2[:len(subs2)], subs[:len(subs)])
subs=subs.replace(subs[:len(subs)],s[i])
i+=1
print("Longest substring in alphabetical order is:",subs2)

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

first_seq=s[0]
break_seq=s[0]
current = s[0]
for i in range(0,len(s)-1):
if s[i]<=s[i+1]:
first_seq = first_seq + s[i+1]
if len(first_seq) > len(current):
current = first_seq
else:
first_seq = s[i+1]
break_seq = first_seq
print("Longest substring in alphabetical order is: ", current)