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Print odd index in same line

I'm trying to complete challenge on HackerRank ( Day 6 : Let's review!) and I only did to print the even numbers on the same line, but I can't print the odd indexes that would be needed to complete the challenge.
This is my code:
word_check = input()
for index, char in enumerate (word_check):
if (index % 2 == 0):
print( char ,end ="" )
This is the most specific task:
Given a string, S , of length N that is indexed from 0 to N -1 , print its even-indexed and odd-indexed characters as space-separated strings on a single line.
Thanks!!!
RavDev

You can use slice notation for indexing the original string:
word_check[::2] + " " + word_check[1::2]
[::2] means "start at the beginning and skip every second element until we reach the end" and [1::2] means "start at the second element and skip every second element until we reach the end". Leaving out either start or stop arguments of the slice implies beginning or end of the sequence respectively. Leaving out the step argument implies a step size of 1.

Slice notation is a better approach, but if you want to use for loop and stick to your approach, you can do in this way:
even =''
odd=''
for index, char in enumerate (word_check):
if (index % 2 == 0):
even += char
else: odd += char
print (even, odd)

I am currently trying to solve the same problem. To get your answers on the same line, initiate two strings: one for even and one for odd. If the character's index is even, add it to the even string and vice versa. Here is my working code so far:
def indexes(word,letter):
result = list()
for i,x in enumerate(word):
if x == letter:
result.append(i)
return result
T = int(input())
if T <= 10 and T>= 1:
for i in range(T):
evenstring = ""
oddstring = ""
lastchar = False
S = input()
if len(S) >= 2 and len(S) <= 10000:
for index, char in enumerate (S):
if (index % 2 == 0):
evenstring += char
else: oddstring += char
if len(indexes(S, char)) > 1:
evenstring.replace(evenstring[evenstring.rfind(char)], '')
oddstring.replace(oddstring[oddstring.rfind(char)], '')
print(evenstring, oddstring)
Your next problem now is trying to remove any reoccurrences of duplicate letters from your final answer (they show up in other test cases)

How to fix a String index out of range exception in Python

There is some issue with my python code. I am making a program that finds the occurrences of the letter A in a word and if that letter is found and the next letter is not the letter A the A is swapped with the next letter.
As an example TAN being TNA but WHOA staying as WHOA
AARDVARK being ARADVRAK
The issue is when I input ABRACADABRA I get a string index out of range exception. Before I had that exception I had the word that prints it as BRACADABRIi'm not sure why if I have to add another loop in my program.
If you guys also have anymore efficient way to run the code then the way I have please let me know!
def scrambleWord(userInput):
count = 0
scramble = ''
while count < len(userInput):
if userInput[count] =='A' and userInput[count+1] != 'A':
scramble+= userInput[count+1] + userInput[count]
count+=2
elif userInput[count] != 'A':
scramble += userInput[count]
count+=1
if count < len(userInput):
scramble += userInput(len(userInput)-1)
return scramble
#if a is found switch the next letter index with a's index
def main():
userInput = input("Enter a word: ")
finish = scrambleWord(userInput.upper())
print(finish)
main()

When you get to the end of the string and it is an 'A' your program is then asking for the next character which is off the end of the string.
Change the loop so it doesn't include the last character:
while count < len(userInput)-1:
if ...

You can modify your code as below:
def scrambleWord(userInput):
count = 0
scramble = ''
while count < len(userInput):
if count < len(userInput)-1 and userInput[count] =='A' and userInput[count+1] != 'A':
scramble+= userInput[count+1] + userInput[count]
count+=2
else:
scramble += userInput[count]
count+=1
return scramble
You are not checking the condition (count < len(userInput)-1) when logic tries to check for A's occurrence and swap with next letter. It throws string index out of range exception.

The issue arises in your code when last character in input is 'A'.
This is because your first if in the loop tries to access 'count + 1' character during last iteration.
And since there's no character at that position, you get index error.
The simplest solution would be to make a separate if condition for the same.
Updated snippet for while loop might look like this -
# while start
while count < len_: # len_ is length of input
if count + 1 >= len_:
break # break outta loop, copy last character
current = inp[count]
next_ = inp[count + 1]
if current == 'A':
op += ( next_ + current) # op is result
count += 1
else:
op += current
# increment counter by 1
count += 1
# rest of the code after while is same
Another small issue in your code is while copying last character ( after loop ends ), you should use [ ] instead of ( ) to refer last character in input string.

Just for fun :
from functools import reduce
def main():
word = input("Enter a word: ").lower()
scramble = reduce((lambda x,y : x[:-1]+y+'A' \
if (x[-1]=='a' and y!=x[-1]) \
else x+y),word)
print(scramble.upper())
main()

how to check repeating Vowels

I'm pretty new to python and I'm having trouble with my
if then else statements and I only get is "no repeating vowels" which mean my rep_vowel is still returning 0
so the program rules are as follows.
if no vowel appears next to itself (e.g. hello), then print:
no vowel repeats
if exactly one vowel is repeated in sequence at least once (e.g. committee) then print a message that indicates which vowel repeats:
only vowel e repeats
if more than one vowel repeats (e.g. green door) then print:
more than one vowel repeats
ignore upper case - lower case differences: assume all the input is always lowercase
answer = input("Enter a string: ")
rep_vowel = 0
i = 0
length_Answer = len(answer)
next_string = 1
curChar = answer[0+rep_vowel]
for i in range(0,length_Answer):
if answer[0 + i] in ["a","e","i","o","u"]:
i =+ 1
next_string = answer[0+i+i]
if next_string == answer:
rep_vowel =+ 1
if rep_vowel == 0:
print("no repeating vowles")
elif rep_vowel > 1:
print("more than 1 repeating vowels")
else:
print ("the letter "+ str(curChar) +" repeats")

You have a few mistakes so i'll try to address several of them:
You do a lot of [0 + something] indexing, which is useless, since 0 + something always equals to something, so yo should just do indexing with [something]
Changing the value of i with i += 1 is bad because you are already increasing it as part of the loop
All you have to do to find a match is simply match the current letter to the next one, if both are the same and they are also vowels, you've found a match.
You are initializing unnecessary variables such as i = 0 only to have them overridden in the next lines
Adding all of those together:
answer = input("Enter a string: ")
vowels = "aeiou"
repeats = [] # this list will hold all repeats of vowels
for i in range(len(answer) - 1): # i'll explain the -1 part at the end
if answer[i] in vowels and answer[i] == answer[i + 1]:
repeats.append(answer[i])
if len(repeats) == 0:
print("no repeating vowles")
elif len(repeats) > 1:
print("more than 1 repeating vowels")
else:
print("the letter " + repeats[0] + " repeats")
This still doesn't take every possible input into account, but it should get you started on a final solution (or perhaps that's enough). For example, input of teest will give the correct result but the input of teeest doesn't (depends on your definition of correct).
About the len(answer-1) range, that's only to make sure we don't go out of bounds when doing answer[i + 1], so we're stopping on the next to last letter instead.

Firstly, you have to indent your code.
to say if (condition) then do print('hello') you write it this way:
if condition:
print('hello')
Secondly, you are using i =+ 1 which is the same as i=1
I think you meant i +=1 which is i = i+1
Finally, I suggest this code:
answer = input("Enter a string: ")
vowel_repeated_count = 0
length_Answer = len(answer)
i=0
while (i <length_Answer-1):
#we check if it's a vowel
if answer[i] in ["a","e","i","o","u"]:
#we check if it's followed by the same vowel
if answer[i+1] == answer[i]:
#increment the vowel_repeated_count
vowel_repeated_count +=1
#we save the vowel for the display
vowel = answer[i]
#we skip the other same repeated vowels
#example: abceeed, we skip the third e
while (answer[i] == vowel and i < length_Answer-1):
i +=1
#we add this incrementation because we're in a while loop
i +=1
if vowel_repeated_count == 0:
print("no repeating vowles")
elif vowel_repeated_count == 1:
print("the letter "+ str(vowel) +" repeats")
else:
print ("more than 1 repeating vowels")

You have some logical errors. It's time consuming to edit that. You can try this, I have modified your code. Hope it will work for you. I have commented beside every important line.
answer = input("Enter a string: ")
is_found = {} #a dictionary that will hold information about how many times a vowel found,initially all are 0
is_found["a"]=0
is_found["e"] = 0
is_found['i']=0
is_found['o']=0
is_found['u']=0
vowels =["a","e","i","o","u"]
for i in range(0,len(answer)):
if answer[i] in vowels:
is_found[answer[i]] = is_found[answer[i]]+1 # if a vowel found then increase its counter
repeated=0 #let 0 repeated vowel
previously_repeated=False #to trace whether there is a previously repeated character found
curChar=None
for key,value in is_found.items(): #iterate over dictionary
if previously_repeated and value>1: #if a vowel found and previously we have another repeated vowel.
repeated=2
elif previously_repeated==False and value>1: # we don't have previously repeated vowel but current vowel is repeated
curChar=key
previously_repeated=True
repeated=1
if repeated== 0:
print("no repeating vowles")
elif repeated> 1:
print("more than 1 repeating vowels")
else:
print ("the letter "+ str(curChar) +" repeats")

There is no need to increment your counter i. In your for loop, it will increment itself each time it goes through the for loop. Also, you need a variable to keep track of how many times the vowel repeats.
answer = input("Enter a string: ")
rep_vowel = 0
length_Answer = len(answer)
vowelList=["a","e","i","o","u"]
vowelRepeated = []
#this will go from i=0 to length_Answer-1
for i in range(length_Answer):
if (answer[i] in vowelList) and (answer[i+1] in vowelList):
if (answer[i] == answer[i+1]):
vowelRepeated.append(answer[i])
repVowel += 1
if rep_vowel==0:
print("no repeating vowels")
elif rep_vowel==1:
print("only one vowel repeated:")
print(vowelRepeated)
else:
print("multiple vowels repeated:")
print(vowelRepeated)

for such counting, I will prefer to use a dictionary to keep the counting number. Your code has been modified for your reference
answer = input("Enter a string: ")
length_Answer = len(answer)
count = dict()
for i in range(length_Answer):
if answer[i] in ["a","e","i","o","u"]:
if answer[i+1] == answer[i]:
if answer[i] in count:
count[answer[i]] += 1
else:
count[answer[i]] = 1
rep_vowel = len(count)
if rep_vowel == 0:
print("no repeating vowles")
elif rep_vowel > 1:
print("more than 1 repeating vowels")
else:
for k in count:
vowel = k
print("the letter " + vowel + " repeats")

You have a few issues with your solution :
1) You never use curChar, i'm guessing you wanted to enter the next_string value into it after the '==' statement.
2) You compare your next_string to answer, this will always be a false statement.
3) Also no need to use [0+i], [i] is good enough
Basically what you want to do is this flow :
1) Read current char
2) Compare to next char
3) If equal put into a different variable
4) If happens again raise a flag
Optional solution :
vowel_list = ["a","e","i","o","u"]
recuring_vowel_boolean_list = [answer[index]==answer[index+1] and answer[index] in vowel_list for index in range(len(answer)-1)]
if not any(recuring_vowel_boolean_list ):
print("no repeating vowels")
elif (recuring_vowel_boolean_list.count(True) > 1):
print("More then 1 repeating vowels")
else:
print("The letter {} repeats".format(answer[recuring_vowel_boolean_list.index(True)]))

How to count specific substrings using slice notation

I want to count the number of occurrences of the substring "bob" within the string s. I do this exercise for an edX Course.
s = 'azcbobobegghakl'
counter = 0
numofiterations = len(s)
position = 0
#loop that goes through the string char by char
for iteration in range(numofiterations):
if s[position] == "b": # search pos. for starting point
if s[position+1:position+2] == "ob": # check if complete
counter += 1
position +=1
print("Number of times bob occurs is: " + str(counter))
However it seems that the s[position+1:position+2] statement is not working properly. How do i adress the two chars behind a "b"?

The second slice index isn't included. It means that s[position+1:position+2] is a single character at position position + 1, and this substring cannot be equal to ob. See a related answer. You need [:position + 3]:
s = 'azcbobobegghakl'
counter = 0
numofiterations = len(s)
position = 0
#loop that goes through the string char by char
for iteration in range(numofiterations - 2):
if s[position] == "b": # search pos. for starting point
if s[position+1:position+3] == "ob": # check if complete
counter += 1
position +=1
print("Number of times bob occurs is: " + str(counter))
# 2

You could use .find with an index:
s = 'azcbobobegghakl'
needle = 'bob'
idx = -1; cnt = 0
while True:
idx = s.find(needle, idx+1)
if idx >= 0:
cnt += 1
else:
break
print("{} was found {} times.".format(needle, cnt))
# bob was found 2 times.

Eric's answer explains perfectly why your approach didn't work (slicing in Python is end-exclusive), but let me propose another option:
s = 'azcbobobegghakl'
substrings = [s[i:] for i in range(0, len(s))]
filtered_s = filter(substrings, lambda s: s.startswith("bob"))
result = len(filtered_s)
or simply
s = 'azcbobobegghakl'
result = sum(1 for ss in [s[i:] for i in range(0, len(s))] if ss.startswith("bob"))

Find the longest substring in alphabetical order

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

s = 'cyqfjhcclkbxpbojgkar'
r = ''
c = ''
for char in s:
if (c == ''):
c = char
elif (c[-1] <= char):
c += char
elif (c[-1] > char):
if (len(r) < len(c)):
r = c
c = char
else:
c = char
if (len(c) > len(r)):
r = c
print(r)

Try changing this:
if len(set(substr)) != (end - start): # found duplicates or EOS
break
if (ord(max(sorted(substr))) - ord(min(sorted(substr))) + 1) == len(substr):
to this:
if len(substr) != (end - start): # found duplicates or EOS
break
if sorted(substr) == list(substr):
That will display ccl for your example input string. The code is simpler because you're trying to solve a simpler problem :-)

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)

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)

In a recursive way, you can import count from itertools
Or define a same method:
def loops( I=0, S=1 ):
n = I
while True:
yield n
n += S
With this method, you can obtain the value of an endpoint, when you create any substring in your anallitic process.
Now looks the anallize method (based on spacegame issue and Mr. Tim Petters suggestion)
def anallize(inStr):
# empty slice (maxStr) to implement
# str native methods
# in the anallize search execution
maxStr = inStr[0:0]
# loop to read the input string (inStr)
for i in range(len(inStr)):
# loop to sort and compare each new substring
# the loop uses the loops method of past
# I = sum of:
# (i) current read index
# (len(maxStr)) current answer length
# and 1
for o in loops(i + len(maxStr) + 1):
# create a new substring (newStr)
# the substring is taked:
# from: index of read loop (i)
# to: index of sort and compare loop (o)
newStr = inStr[i:o]
if len(newStr) != (o - i):# detect and found duplicates
break
if sorted(newStr) == list(newStr):# compares if sorted string is equal to listed string
# if success, the substring of sort and compare is assigned as answer
maxStr = newStr
# return the string recovered as longest substring
return maxStr
Finally, for test or execution pourposes:
# for execution pourposes of the exercise:
s = "azcbobobegghakl"
print "Longest substring in alphabetical order is: " + anallize( s )
The great piece of this job started by: spacegame and attended by Mr. Tim Petters, is in the use of the native str methods and the reusability of the code.
The answer is:
Longest substring in alphabetical order is: ccl

In Python character comparison is easy compared to java script where the ASCII values have to be compared. According to python
a>b gives a Boolean False and b>a gives a Boolean True
Using this the longest sub string in alphabetical order can be found by using the following algorithm :
def comp(a,b):
if a<=b:
return True
else:
return False
s = raw_input("Enter the required sting: ")
final = []
nIndex = 0
temp = []
for i in range(nIndex, len(s)-1):
res = comp(s[i], s[i+1])
if res == True:
if temp == []:
#print i
temp.append(s[i])
temp.append(s[i+1])
else:
temp.append(s[i+1])
final.append(temp)
else:
if temp == []:
#print i
temp.append(s[i])
final.append(temp)
temp = []
lengths = []
for el in final:
lengths.append(len(el))
print lengths
print final
lngStr = ''.join(final[lengths.index(max(lengths))])
print "Longest substring in alphabetical order is: " + lngStr

Use list and max function to reduce the code drastically.
actual_string = 'azcbobobegghakl'
strlist = []
i = 0
while i < len(actual_string)-1:
substr = ''
while actial_string[i + 1] > actual_string[i] :
substr += actual_string[i]
i += 1
if i > len(actual_string)-2:
break
substr += actual-string[i]
i += 1
strlist.append(subst)
print(max(strlist, key=len))

Wow, some really impressing code snippets here...
I want to add my solution, as I think it's quite clean:
s = 'cyqfjhcclkbxpbojgkar'
res = ''
tmp = ''
for i in range(len(s)):
tmp += s[i]
if len(tmp) > len(res):
res = tmp
if i > len(s)-2:
break
if s[i] > s[i+1]:
tmp = ''
print("Longest substring in alphabetical order is: {}".format(res))

Without using a library, but using a function ord() which returns ascii value for a character.
Assumption: input will be in lowercase, and no special characters are used
s = 'azcbobobegghakl'
longest = ''
for i in range(len(s)):
temp_longest=s[i]
for j in range(i+1,len(s)):
if ord(s[i])<=ord(s[j]):
temp_longest+=s[j]
i+=1
else:
break
if len(temp_longest)>len(longest):
longest = temp_longest
print(longest)

Slightly different implementation, building up a list of all substrings in alphabetical order and returning the longest one:
def longest_substring(s):
in_orders = ['' for i in range(len(s))]
index = 0
for i in range(len(s)):
if (i == len(s) - 1 and s[i] >= s[i - 1]) or s[i] <= s[i + 1]:
in_orders[index] += s[i]
else:
in_orders[index] += s[i]
index += 1
return max(in_orders, key=len)

s = "azcbobobegghakl"
ls = ""
for i in range(0, len(s)-1):
b = ""
ss = ""
j = 2
while j < len(s):
ss = s[i:i+j]
b = sorted(ss)
str1 = ''.join(b)
j += 1
if str1 == ss:
ks = ss
else:
break
if len(ks) > len(ls):
ls = ks
print("The Longest substring in alphabetical order is "+ls)

This worked for me
s = 'cyqfjhcclkbxpbojgkar'
lstring = s[0]
slen = 1
for i in range(len(s)):
for j in range(i,len(s)-1):
if s[j+1] >= s[j]:
if (j+1)-i+1 > slen:
lstring = s[i:(j+1)+1]
slen = (j+1)-i+1
else:
break
print("Longest substring in alphabetical order is: " + lstring)
Output: Longest substring in alphabetical order is: ccl

input_str = "cyqfjhcclkbxpbojgkar"
length = len(input_str) # length of the input string
iter = 0
result_str = '' # contains latest processed sub string
longest = '' # contains longest sub string alphabetic order
while length > 1: # loop till all char processed from string
count = 1
key = input_str[iter] #set last checked char as key
result_str += key # start of the new sub string
for i in range(iter+1, len(input_str)): # discard processed char to set new range
length -= 1
if(key <= input_str[i]): # check the char is in alphabetic order
key = input_str[i]
result_str += key # concatenate the char to result_str
count += 1
else:
if(len(longest) < len(result_str)): # check result and longest str length
longest = result_str # if yes set longest to result
result_str = '' # re initiate result_str for new sub string
iter += count # update iter value to point the index of last processed char
break
if length is 1: # check for the last iteration of while loop
if(len(longest) < len(result_str)):
longest = result_str
print(longest);

finding the longest substring in alphabetical order in Python
in python shell 'a' < 'b' or 'a' <= 'a' is True
result = ''
temp = ''
for char in s:
if (not temp or temp[-1] <= char):
temp += char
elif (temp[-1] > char):
if (len(result) < len(temp)):
result = temp
temp = char
if (len(temp) > len(result)):
result = temp
print('Longest substring in alphabetical order is:', result)

s=input()
temp=s[0]
output=s[0]
for i in range(len(s)-1):
if s[i]<=s[i+1]:
temp=temp+s[i+1]
if len(temp)>len(output):
output=temp
else:
temp=s[i+1]
print('Longest substring in alphabetic order is:' + output)

I had similar question on one of the tests on EDX online something. Spent 20 minutes brainstorming and couldn't find solution. But the answer got to me. And it is very simple. The thing that stopped me on other solutions - the cursor should not stop or have unique value so to say if we have the edx string s = 'azcbobobegghakl' it should output - 'beggh' not 'begh'(unique set) or 'kl'(as per the longest identical to alphabet string). Here is my answer and it works
n=0
for i in range(1,len(s)):
if s[n:i]==''.join(sorted(s[n:i])):
longest_try=s[n:i]
else:
n+=1

In some cases, input is in mixed characters like "Hello" or "HelloWorld"
**Condition 1:**order determination is case insensitive, i.e. the string "Ab" is considered to be in alphabetical order.
**Condition 2:**You can assume that the input will not have a string where the number of possible consecutive sub-strings in alphabetical order is 0. i.e. the input will not have a string like " zxec ".
string ="HelloWorld"
s=string.lower()
r = ''
c = ''
last=''
for char in s:
if (c == ''):
c = char
elif (c[-1] <= char):
c += char
elif (c[-1] > char):
if (len(r) < len(c)):
r = c
c = char
else:
c = char
if (len(c) > len(r)):
r = c
for i in r:
if i in string:
last=last+i
else:
last=last+i.upper()
if len(r)==1:
print(0)
else:
print(last)
Out:elloW

```python
s = "cyqfjhcclkbxpbojgkar" # change this to any word
word, temp = "", s[0] # temp = s[0] for fence post problem
for i in range(1, len(s)): # starting from 1 not zero because we already add first char
x = temp[-1] # last word in var temp
y = s[i] # index in for-loop
if x <= y:
temp += s[i]
elif x > y:
if len(temp) > len(word): #storing longest substring so we can use temp for make another substring
word = temp
temp = s[i] #reseting var temp with last char in loop
if len(temp) > len(word):
word = temp
print("Longest substring in alphabetical order is:", word)
```
My code store longest substring at the moment in variable temp, then compare every string index in for-loop with last char in temp (temp[-1]) if index higher or same with (temp[-1]) then add that char from index in temp. If index lower than (temp[-1]) checking variable word and temp which one have longest substring, after that reset variable temp so we can make another substring until last char in strings.

s = 'cyqfjhcclkbxpbojgkar'
long_sub = '' #longest substring
temp = '' # temporarily hold current substr
if len(s) == 1: # if only one character
long_sub = s
else:
for i in range(len(s) - 1):
index = i
temp = s[index]
while index < len(s) - 1:
if s[index] <= s[index + 1]:
temp += s[index + 1]
else:
break
index += 1
if len(temp) > len(long_sub):
long_sub = temp
temp = ''
print(long_sub)

For comprehensibility, I also add this code snippet based on regular expressions. It's hard-coded and seems clunky. On the other hand, it seems to be the shortest and easiest answer to this problem. And it's also among the most efficient in terms of runtime complexity (see graph).
import re
def longest_substring(s):
substrings = re.findall('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*', s)
return max(substrings, key=len)
(Unfortunately, I'm not allowed to paste a graph here as a "newbie".)
Source + Explanation + Graph: https://blog.finxter.com/python-how-to-find-the-longest-substring-in-alphabetical-order/

Another way:
s = input("Please enter a sentence: ")
count = 0
maxcount = 0
result = 0
for char in range(len(s)-1):
if(s[char]<=s[char+1]):
count += 1
if(count > maxcount):
maxcount = count
result = char + 1
else:
count = 0
startposition = result - maxcount
print("Longest substring in alphabetical order is: ", s[startposition:result+1])