Related
Hope you all are doing well in these times.
here's my code:
def ab(n):
first = 0
last = -1
endprod = n[first] + n[last]
endprod2 = n[first+1] + n[last-1]
endprod3 = n[first+2] + n[last-2]
endprod4 = n[first+3] + n[last-3]
endprod5 = n[first+4] + n[last-4]
endprod100 = endprod[::-1] + endprod2[::-1] + endprod3[::-1]+ endprod4[::-1]+ endprod5[::-1]
return endprod100
I was able do to it, however mine isn't a loop. Is there a way to convert my code into a for loop. So, increment by 1 and decrement by 1.
Thanks,
Try this:
def ab(n):
result = ''
for j in range(len(n) // 2):
result += n[-j-1] + n[j]
if len(n) % 2 == 1:
result += n[len(n) // 2]
return result
You also need the part
if len(n) % 2 == 1:
result += n[len(n) // 2]
because your input string might have an odd number of characters
Examples:
>>> ab('0123456789')
'9081726354'
>>> ab('01234956789')
'90817263549'
If you want to reuse your original logic:
def ab(n):
result = ''
first = 0
last = -1
for j in range(len(n) // 2):
result += n[last-j] + n[first+j]
if len(n) % 2 == 1:
result += n[len(n) // 2]
return result
You could also recurse it:
def ab(s):
if len(s)>2:
return s[-1]+s[0]+ab(s[1:-1])
else:
return s
But the last part of Riccardo's answer fits your question more closely.
you need to split your string for your loop, means first you broke your string to half then build your string, you could use zip to iterate over multiple iteratable. something like this:
def ab(s):
out = ""
for v0,v1 in zip(s[:len(s)//2 -1 :-1], s[:len(s)//2 ]):
out += v0 + v1
return out
the better version you should write without loop.
like this:
out = "".join(map(lambda x: "".join(x), zip(s[:len(s)//2 -1 :-1], s[:len(s)//2 ])))
There are already a lot of good answers that are probably clearer, easier to read and much better suited for learning purposes than what I'm about to write. BUT... something I wanted to bring up (maybe just telling myself, because I tend to forget it) is that sometimes destroying the original argument can facilitate this sort of things.
In this case, you could keep "poping" left and right, shrinking your string until you exhaust it
def swap_destr(my_str):
result = ""
while len(my_str) > 1:
result += my_str[-1] # Last char
result += my_str[0] # First char
my_str = my_str[1:-1] # Shrink it!
return result + my_str # + my_str just in case there 1 char left
print(swap_destr("0123456789"))
print(swap_destr("123456789"))
# Outputs:
# 9081726354
# 918273645
This is also a good problem to see (and play with) recursion:
def swap_recur(my_str):
if len(my_str) <= 1:
return my_str
return my_str[-1] + my_str[0] + swap_recur(my_str[1:-1])
print(swap_recur("0123456789"))
print(swap_recur("123456789"))
# Outputs:
# 9081726354
# 918273645
EDIT:
Thanks for fixing it! Unfortunatelly, it messed up the logic. I'll explain what this program does. It's a solution to a task about playing cards trick. There are N cards on the table. First and Second are numbers on the front and back of the cards. The trick can only be done, if the visible numbers are in non-decreasing order. Someone from audience can come and swap places of cards. M represents how many cards will be swapped places. A and B represent which cards will be swapped. Magician can flip any number of cards to see the other side. The program must tell, if the magician can do the trick.
from collections import namedtuple
Pair = namedtuple("Pair", ["first", "second"])
pairs = []
with open('data.txt', 'r') as data, open('results.txt', 'w') as results:
n = data.readline()
n = int(n)
for _ in range(n):
first, second = (int(x) for x in data.readline().split(':'))
first, second = sorted((first, second))
pairs.append(Pair(first, second)) # add to the list by appending
m = data.readline()
m = int(m)
for _ in range(m):
a, b = (int(x) for x in data.readline().split('-'))
a -= 1
b -= 1
temp = pairs[a]
pairs[a] = pairs[b]
pairs[b] = temp
p = -1e-9
ok = True
for k in range(0, n):
if pairs[k].first >= p:
p = pairs[k].first
elif pairs[k].second >= p:
p = pairs[k].second
else:
ok = False
break
if ok:
results.write("YES\n")
else:
results.write("NO\n")
data:
4
2:5
3:4
6:3
2:7
2
3-4
1-3
results:
YES
YES
YES
YES
YES
YES
YES
What should be in results:
NO
YES
The code is full of bugs: you should write and test it incrementally instead of all at once. It seems that you started using readlines (which is a good way of managing this kind of work) but you kept the rest of the code in a reading one by one style. If you used readlines, the line for i, line in enumerate(data): should be changed to for i, line in enumerate(lines):.
Anyway, here is a corrected version with some explanation. I hope I did not mess with the logic.
from collections import namedtuple
Pair = namedtuple("Pair", ["first", "second"])
# The following line created a huge list of "Pairs" types, not instances
# pairs = [Pair] * (2*200*1000+1)
pairs = []
with open('data.txt', 'r') as data, open('results.txt', 'w') as results:
n = data.readline()
n = int(n)
# removing the reading of all data...
# lines = data.readlines()
# m = lines[n]
# removed bad for: for i, line in enumerate(data):
for _ in range(n): # you don't need the index
first, second = (int(x) for x in data.readline().split(':'))
# removed unnecessary recasting to int
# first = int(first)
# second = int(second)
# changed the swapping to a more elegant way
first, second = sorted((first, second))
pairs.append(Pair(first, second)) # we add to the list by appending
# removed unnecessary for: once you read all the first and seconds,
# you reached M
m = data.readline()
m = int(m)
# you don't need the index... indeed you don't need to count (you can read
# to the end of file, unless it is malformed)
for _ in range(m):
a, b = (int(x) for x in data.readline().split('-'))
# removed unnecessary recasting to int
# a = int(a)
# b = int(b)
a -= 1
b -= 1
temp = pairs[a]
pairs[a] = pairs[b]
pairs[b] = temp
p = -1e-9
ok = True
for k in range(0, n):
if pairs[k].first >= p:
p = pairs[k].first
elif pairs[k].second >= p:
p = pairs[k].second
else:
ok = False
break
if ok:
results.write("YES\n")
else:
results.write("NO\n")
Response previous to edition
range(1, 1) is empty, so this part of the code:
for i in range (1, 1):
n = data.readline()
n = int(n)
does not define n, at when execution gets to line 12 you get an error.
You can remove the for statement, changing those three lines to:
n = data.readline()
n = int(n)
Let's say you have a list which only has two types of values and goes something like ['t','r','r','r','t','t','r','r','t'] and you want to find the length of the smallest sequence number of 'r's which have 't's at both ends.
In this case the smallest sequence of 'r' has a length of 2, because there is first t,r,r,r,t and then t,r,r,t, and the latter has the smallest number of 'r's in a row surrounded by 't' and the number of 'r's is 2.
How would I code for finding that number?
This is from a problem of trying of going to a play with your friend, and you want to sit as close as possible with your friend, so you are trying to find the smallest amount of taken seats in between two free seats at a play. "#" is a taken seat and a "." is a free seat. you are given the amount of seats, and the seating arrangement (free seats and taken seats), and they are all in one line.
An example of an input is:
5
#.##.
where there are two taken seats(##) in between two free seats.
Here is my code which is not working for inputs that I don't know, but working for inputs I throw at it.
import sys
seats = int(input())
configuration = input()
seatsArray = []
betweenSeats = 1
betweenSeatsMin = 1
checked = 0
theArray = []
dotCount = 0
for i in configuration:
seatsArray.append(i)
for i in range(len(seatsArray)):
if i == len(seatsArray) - 1:
break
if seatsArray[i] == "." and seatsArray[i+1] == ".":
print(0)
sys.exit()
for i in range(0,len(seatsArray)):
if i > 0:
if checked == seats:
break
checked += 1
if seatsArray[i] == "#":
if i > 0:
if seatsArray[i-1] == "#":
betweenSeats += 1
if seatsArray[i] == ".":
dotCount += 1
if dotCount > 1:
theArray.append(betweenSeats)
betweenSeats = 1
theArray = sorted(theArray)
if theArray.count(1) > 0:
theArray.remove(1)
theArray = list(dict.fromkeys(theArray))
print(theArray[0])
This is a noob and a !optimal approach to your problem using a counter for the minimum and maximum sequence where ew compare both and return the minimum.
''' create a funciton that
will find min sequence
of target char
in a list'''
def finder(a, target):
max_counter = 0
min_counter = 0
''' iterate through our list
and if the element is the target
increase our max counter by 1
'''
for i in x:
if i == target:
max_counter += 1
'''min here is 0
so it will always be less
so we overwrite it's value
with the value of max_counter'''
if min_counter < max_counter:
min_counter = max_counter
'''at last iteration max counter will be less than min counter
so we overwrite it'''
if max_counter < min_counter:
min_counter = max_counter
else:
max_counter = 0
return min_counter
x = ['t','r','r','r','t','t','r','r','t','t','t','r','t']
y = 'r'
print(finder(x,y))
Create a string from list and then search for pattern required and then count r in the found matches and then take min of it
Code:
import re
lst = ['t','r','r','r','t','t','r','r','t']
text = ''.join(lst)
pattern = '(?<=t)r+(?=t)'
smallest_r_seq = min(match.group().count('r') for match in re.finditer(pattern, text))
print(smallest_r_seq)
Output:
2
I am doing the Project Euler #67 in Python. My program, which worked for Project 18, does not work for Project 67.
Code (excludes the opening of the file and the processing of information):
for i in range(len(temp)):
list1 = temp[i]
try:
list2 = temp[i+1]
trynum1 = list1[lastinput] + max(list2[lastinput],list2[lastinput+1])
try:
trynum2 = list1[lastinput+1] + max(list2[lastinput+1],list2[lastinput+2])
if trynum1 > trynum2:
outputlist.append(list1[lastinput])
else:
outputlist.append(list1[lastinput+1])
lastinput += 1
except IndexError:
outputlist.append(list1[0])
except IndexError:
if list1[lastinput] > list1[lastinput+1]:
outputlist.append(list1[lastinput])
else:
outputlist.append(list1[lastinput+1])
Variables:
temp is the triangle of integers
outputlist is a list which stores the numbers chosen by the program
I know the answer is 7273, but my program finds 6542. I cannot find an error which causes the situation. Please may you help me on it.
Logic
My approach to this program is to find one number (list1[lastinput]) and add it up with the larger number of the two below it (trynum1), compare with the number to the right of the first number (list1[lastinput+1]), adding the larger number of two below it (trynum2). I append the larger one to the output list.
This approach is logically flawed. When you're in row 1, you don't have enough information to know whether moving right or left will lead you to the largest sum, not with only a 2-row lookahead. You would need to look all the way to the bottom to ensure getting the best path.
As others have suggested, start at the bottom and work up. Remember, you don't need the entire path, just the sum. At each node, add the amount of the better of the two available paths (that's the score you get in taking that node to the bottom). When you get back to the top, temp[0][0], that number should be your final answer.
I thought day and night about problem 18 and I solved it, the same way I solved this one.
P.S. 100_triangle.txt is without 1st string '59'.
# Maximum path sum II
import time
def e67():
start = time.time()
f=open("100_triangle.txt")
summ=[59]
for s in f:
slst=s.split()
lst=[int(item) for item in slst]
for i in range(len(lst)):
if i==0:
lst[i]+=summ[i]
elif i==len(lst)-1:
lst[i]+=summ[i-1]
elif (lst[i]+summ[i-1])>(lst[i]+summ[i]):
lst[i]+=summ[i-1]
else:
lst[i]+=summ[i]
summ=lst
end = time.time() - start
print("Runtime =", end)
f.close()
return max(summ)
print(e67()) #7273
Though starting from the bottom is more efficient, I wanted to see if I could implement Dijkstra's algorithm on this one; it works well and only takes a few seconds (didn't time it precisely):
from math import inf
f = open("p067_triangle.txt", "r")
tpyramid = f.read().splitlines()
f.close()
n = len(tpyramid)
pyramid = [[100 - int(tpyramid[i].split()[j]) for j in range(i+1)] for i in range(n)]
paths = [[inf for j in range(i+1)] for i in range(n)]
paths[0][0] = pyramid[0][0]
def mini_index(pyr):
m = inf
for i in range(n):
mr = min([i for i in pyr[i] if i >= 0]+[inf])
if mr < m:
m, a, b = mr, i, pyr[i].index(mr)
return m, a, b
counter = 0
omega = inf
while counter < n*(n+1)/2:
min_weight, i, j = mini_index(paths)
if i != n-1:
paths[i+1][j] = min( paths[i+1][j], min_weight + pyramid[i+1][j])
paths[i+1][j+1] = min( paths[i+1][j+1], min_weight + pyramid[i+1][j+1])
else:
omega = min(omega, min_weight)
paths[i][j] = -1
counter += 1
print(100*n - omega)
Here is my solution. Indeed you have to take the bottom - up approach.
Result confirmed with PE. Thanks!
def get_triangle(listLink):
triangle = [[int(number) for number in row.split()] for row in open(listLink)]
return triangle
listOfLists = get_triangle('D:\\Development\\triangle.txt')
for i in range(len(listOfLists) - 2, -1, -1):
for j in range(len(listOfLists[i])):
listOfLists[i][j] += max(listOfLists[i+1][j], listOfLists[i+1][j+1])
print(listOfLists[0][0])
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)