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Input is xyz = 'aaabbbaaa', I want output as 3a3b3a
xyz = 'aaabbbaaa'
p = xyz[0]
i = 0
out = {}
while i < len(xyz):
if p == xyz[i]:
if xyz[i] not in out:
out[xyz[i]] = []
out[xyz[i]].append(xyz[i])
else:
p = xyz[i]
i += 1
print(out)
Help me, How can i achieve this??
This is likely the simplest method and easiest to understand.
Create a tally variable and increment it when you see repeating characters, then when you see a non repeating character write the previous character and the tally to a string and start the tally back to 1.... repeat until string ends
xyz = 'aaabbbaaa'
tally = 1
string = ''
prev = xyz[0]
for char in xyz[1:]:
if char == prev:
tally += 1
else:
string += str(tally) + prev
prev = char
tally = 1
string += str(tally) + prev
print(string) # 3a3b3a
what result do you expect to get if the string has single characters? suppose we should just skip a single character:
from re import sub
s = 'aabbbcaaabc'
sub(r'(\w)\1*',lambda m: f"{l if (l:=len(m[0]))>1 else ''}{m[1]}",s)
>>>
# '2a3bc3abc'
For example, if the string was "pwwkew", the longest substring without repeating characters would be "wke".
def longest_non_repeating_substring():
count = 0
current_longest = 0
consideration = []
possible_longest = []
while count > len(string):
current_char = string[count:count+1]
consideration.append(current_char)
for i in range(len(consideration)):
if current_char == consideration[i]:
possible_longest.append(current_longest)
current_longest = 0
del consideration[:]
consideration.append(current_char)
current_longest += 1
count += 1
return max(possible_longest)
longest_non_repeating_substring("pwwkew")
So, I iterate through the string one character at a time. At each character, I add it to my array consideration, and check if that current character already exists there. If it does, I reset the array consideration, make current longest zero, and leave in the current character in the array. If the current character isnt there then I increase the counter and the length of the current longest. It seems very logical to me, but this is wrong. Anyone know what gives?
I think the given code looks a little bit complicated. You could try to use something similar to this:
def longest_non_repeating_substring(string):
count = 0
current_longest = []
current_candidate = []
# Iterate over all characters in the string
while count < len(string):
current_char = string[count:count+1]
# Check if we have a duplicate
if current_char in current_candidate:
# if so create a new candidate with the substring after the duplicate
index_duplicate = current_candidate.index(current_char)
current_candidate = current_candidate[index_duplicate+1:]
# Append current character to candidate
current_candidate.append(current_char)
# Check if candidate is longer than what we had before; if so make candidate current longest
if len(current_candidate) > len(current_longest):
current_longest = current_candidate
count = count +1
# Convert list of current_longest to string
current_longest_str = ''.join(current_longest)
return current_longest_str
string = 'pwwkew'
longest = longest_non_repeating_substring(string)
print(longest) # Output: wke
I see that others already pointed you in the right direction. I propose the following solution:
def longest_non_repeating_chars_substring(string: str):
longest_substring = None
current_substring = ''
for i, char in enumerate(string):
if i > 0:
if char != string[i - 1] or current_substring == '':
current_substring += char
else:
longest_substring = current_substring
current_substring = char
else:
current_substring += char
if len(current_substring) > len(longest_substring):
return current_substring
return longest_substring
Let me know of your thoughts!
I need a Python function which gives reversed string with the following conditions.
$ position should not change in the reversed string.
Should not use Python built-in functions.
Function should be an efficient one.
Example : 'pytho$n'
Result : 'nohty$p'
I have already tried with this code:
list = "$asdasdas"
list1 = []
position = ''
for index, i in enumerate(list):
if i == '$':
position = index
elif i != '$':
list1.append(i)
reverse = []
for index, j in enumerate( list1[::-1] ):
if index == position:
reverse.append( '$' )
reverse.append(j)
print reverse
Thanks in advance.
Recognise that it's a variation on the partitioning step of the Quicksort algorithm, using two pointers (array indices) thus:
data = list("foo$barbaz$$")
i, j = 0, len(data) - 1
while i < j:
while i < j and data[i] == "$": i += 1
while i < j and data[j] == "$": j -= 1
data[i], data[j] = data[j], data[i]
i, j = i + 1, j - 1
"".join(data)
'zab$raboof$$'
P.S. it's a travesty to write this in Python!
A Pythonic solution could look like this:
def merge(template, data):
for c in template:
yield c if c == "$" else next(data)
data = "foo$barbaz$$"
"".join(merge(data, reversed([c for c in data if c != "$"])))
'zab$raboof$$'
Wrote this without using any inbuilt functions. Hope it fulfils your criteria -
string = "zytho$n"
def reverse(string):
string_new = string[::-1]
i = 0
position = 0
position_new = 0
for char in string:
if char=="$":
position = i
break
else:
i = i + 1
j = 0
for char in string_new:
if char=="$":
position_new = i
break
else:
j = j + 1
final_string = string_new[:position_new]+string_new[position_new+1:position+1]+"$"+string_new[position+1:]
return(final_string)
string_new = reverse(string)
print(string_new)
The output of this is-
nohty$x
To explain the code to you, first I used [::-1], which is just taking the last position of the string and moving forward so as to reverse the string. Then I found the position of the $ in both the new and the old string. I found the position in the form of an array, in case you have more than one $ present. However, I took for granted that you have just one $ present, and so took the [0] index of the array. Next I stitched back the string using four things - The part of the new string upto the $ sign, the part of the new string from after the dollar sign to the position of the $ sign in the old string, then the $ sign and after that the rest of the new string.
My code seems to work with shorter strings, but inexplicably to me gets stuck on others. The function of this is to replace characters with digits, and I have it print out the new string after each part is replaced. Any help you can give me is appreciated, thanks!
By the way, I did look at the similar questions on this and they did not answer my particular question, please don't remove my question.
possibleChars = ['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','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','1','2','3','4','5','6','7','8','9','0',' ',',','.','?','!','/','\\','[',']','{','}',
'|','<','>',';',':','+','=','-','_','(',')','#','#','$','%','^','&','*','~','`'] #0-92
possibleCharsToDigit = ['1','2','3','4','5','6','7','8','9','0','1','2','3','4','5','6','7','8','9','0','1','2','3',
'4','5','6','7','8','9','0','1','2','3','4','5','6','7','8','9','0','1','2','3','4','5','6','7','8',
'9','0','1','2','3','4','5','6','7','8','9','0','1','2','3','4','5','6','7','8','9','0','1','2','3',
'4','5','6','7','8','9','0','1','2','3','4','5','6','7','8','9','0','1','2','3'] #0-92
password = "How is your day today?"
def passwordToDigit(passToConvert):
passLen = len(passToConvert) #puts the length of the password in a variable
i = 0 #i is the selected character in the password
j = 0 #j is the selected possible char, i.e. '0' is 'A' in possibleChars or '1' in possibleCharsToDigit
while i < passLen:
if passToConvert[i] == possibleChars[j]:
passToConvert = passToConvert[0:i] + possibleCharsToDigit[j] + passToConvert[i + 1:]
i += 1
print passToConvert
else:
j += 1
print passToConvert
passwordToDigit(password)
When you are incrementing j variable inside the while loop, notice that when j gets bigger than the length of possibleCharsToDigit list then you are trying to access its element with index out of bounds.
you should set j = 0 in the if passToConvert[i] == possibleChars[j] clause:
def passwordToDigit(passToConvert):
passLen = len(passToConvert) #puts the length of the password in a variable
i = 0 #i is the selected character in the password
j = 0 #j is the selected possible char, i.e. '0' is 'A' in possibleChars or '1' in possibleCharsToDigit
while i < passLen:
if passToConvert[i] == possibleChars[j]:
passToConvert = passToConvert[0:i] + possibleCharsToDigit[j] + passToConvert[i + 1:]
i += 1
j = 0
print passToConvert
else:
j += 1
print passToConvert
As you increment j within your while loop, without ever resetting it, each time you successfully match a character and move onto the next one. This will cause your code to fail as soon as you have a character earlier in possibleChars than a previous one.
To illustrate:
passwordToDigit('ABCDEFGHIJKLMNOP') #will work correctly
passwordToDigit('BA') #will fail with IndexError
Quick Solution
The quickest solution would be to reset the j index when you find a match.
ie
# ...
if passToConvert[i] == possibleChars[j]:
passToConvert = passToConvert[0:i] + possibleCharsToDigit[j] + passToConvert[i + 1:]
i += 1
print passToConvert
j = 0 #Reset tje j index to start searching from beginning
else:
#...
Dict Solution
You could also spend some time refactoring your code to use a dict to map characters to digits as in:
import string
charopts = string.ascii_uppercase + string.ascii_lowercase + string.digits[1:] + r'0 ,.?!/\[]{}|<>;:+=-_()##$%^&*~`'
char2dig = dict((k,str((i+1)%10)) for i,k in enumerate(charopts))
def passwordToDigitDic(passToConvert):
newpass = ''
for c in passToConvert:
newpass += char2dig[c]
print(newpass + passToConvert[len(newpass):])
passwordToDigitDic('ABCDEFGH')
passwordToDigitDic('HGEFBCA')
Note, if you are ever interested in doing the translation in one go as opposed to step by step with prints, look into the string.translate function.
When i=passLen-1, then you are trying to access passToConvert[i+1], which is out of the range of passToConvert. Hence you are getting this error. Try this:
if passToConvert[i] == possibleChars[j]:
if i<passLen-1:
passToConvert = passToConvert[0:i] + possibleCharsToDigit[j] + passToConvert[i + 1:]
else:
passToConvert = passToConvert[0:i] + possibleCharsToDigit[j]
i += 1
print passToConvert
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)