Develop Reference

Python Inserting a string

I need to insert a string (character by character) into another string at every 3rd position
For example:- string_1:-wwwaabkccgkll
String_2:- toadhp
Now I need to insert string2 char by char into string1 at every third position
So the output must be wwtaaobkaccdgkhllp
Need in Python.. even Java is ok
So i tried this
Test_str="hiimdumbiknow"
challenge="toadh"
new_st=challenge [k]
Last=list(test_str)
K=0
For i in range(Len(test_str)):
if(i%3==0):
last.insert(i,new_st)
K+=1
and the output i get
thitimtdutmbtiknow

You can split test_str into sub-strings to length 2, and then iterate merging them with challenge:
def concat3(test_str, challenge):
chunks = [test_str[i:i+2] for i in range(0,len(test_str),2)]
result = []
i = j = 0
while i<len(chunks) or j<len(challenge):
if i<len(chunks):
result.append(chunks[i])
i += 1
if j<len(challenge):
result.append(challenge[j])
j += 1
return ''.join(result)
test_str = "hiimdumbiknow"
challenge = "toadh"
print(concat3(test_str, challenge))
# hitimoduambdikhnow
This method works even if the lengths of test_str and challenge are mismatching. (The remaining characters in the longest string will be appended at the end.)

You can split Test_str in to groups of two letters and then re-join with each letter from challenge in between as follows;
import itertools
print(''.join(f'{two}{letter}' for two, letter in itertools.zip_longest([Test_str[i:i+2] for i in range(0,len(Test_str),2)], challenge, fillvalue='')))
Output:
hitimoduambdikhnow
*edited to split in to groups of two rather than three as originally posted

you can try this, make an iter above the second string and iterate over the first one and select which character should be part of the final string according the position
def add3(s1, s2):
def n():
try:
k = iter(s2)
for i,j in enumerate(s1):
yield (j if (i==0 or (i+1)%3) else next(k))
except:
try:
yield s1[i+1:]
except:
pass
return ''.join(n())

def insertstring(test_str,challenge):
result = ''
x = [x for x in test_str]
y = [y for y in challenge]
j = 0
for i in range(len(x)):
if i % 2 != 0 or i == 0:
result += x[i]
else:
if j < 5:
result += y[j]
result += x[i]
j += 1
get_last_element = x[-1]
return result + get_last_element
print(insertstring(test_str,challenge))
#output: hitimoduambdikhnow

Python: how to replace a substring with a number of its occurences?

Let's say I have a string presented in the following fashion:
st = 'abbbccccaaaAAbccc'
The task is to encode it so that single characters are followed by a number of their occurences:
st = 'a1b3c4a3A2b1c3'
I know one possible solution but it's too bulky and primitive.
s = str(input())
l = len(s)-1
c = 1
t = ''
if len(s)==1:
t = t +s+str(c)
else:
for i in range(0,l):
if s[i]==s[i+1]:
c +=1
elif s[i]!=s[i+1]:
t = t + s[i]+str(c)
c = 1
for j in range(l,l+1):
if s[-1]==s[-2]:
t = t +s[j]+str(c)
elif s[-1]!=s[-2]:
t = t +s[j]+str(c)
c = 1
print(t)
Is there any way to solve this shortly and elegantly?
P.S: I'm an unexperienced Python user and a new StackOverflow member, so I beg my pardon if the question is asked incorrectly.

Take advantage of the standard library:
from itertools import groupby
st = "abbbccccaaaAAbccc"
print("".join("{}{}".format(key, len(list(group))) for key, group in groupby(st)))
Output:
a1b3c4a3A2b1c3
>>>

just loop through and count. There are more graceful snippets but this will get the job done and is clear:
count = 1
char = st[0]
new_st = []
for c in st[1:]:
if c == char:
count += 1
else:
new_st.append(char + str(count))
char = c
count = 1
new_st.append(char + str(count))
s2= "".join(new_st)
print(s2) # 'a1b3c4a3A2b1c3'
If you want a fancy recursive solution:
def compose(s):
if not s:
return ""
count = 1
for char in s[1:]:
if s[0] != char:
break
count += 1
return s[0] + str(count) + compose(s[count:])

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)

Longest substring without repeating characters in python

This is a pretty standard interview question. Find the longest substring without repeating characters. Here are the test cases,
abcabcbb -> 3
bbbbb -> 1
pwwkew -> 3
bpfbhmipx -> 7
tmmzuxt -> 5
Here's my code which uses a pretty simple approach with 2 pointers.
def lengthOfLongestSubstring(s):
checklist = {}
starting_index_of_current_substring = 0
length_of_longest_substring = 0
for i, v in enumerate(s):
if v in checklist:
starting_index_of_current_substring = checklist[v] + 1
else:
length_of_current_substring = i - starting_index_of_current_substring + 1
length_of_longest_substring = max(length_of_current_substring, length_of_longest_substring)
checklist[v] = i
return length_of_longest_substring
My code passes all the test cases except the last one (actual 4, expected 5). Can someone help me modify the code to take care of the last test case. I don't wish to reinvent the algorithm.

Here is a simple tweak in your code with 2 pointers to find the longest sub-string without repeating characters.
Change in your code is instead of calculating the length of longest substring when v is not present in checklist, I am calculating length of longest substring for all cases.
def lengthOfLongestSubstring(s):
checklist = {}
starting_index_of_current_substring = 0
length_of_longest_substring = 0
for i, v in enumerate(s):
if v in checklist:
starting_index_of_current_substring = max(starting_index_of_current_substring, checklist[v] + 1)
checklist[v] = i
length_of_longest_substring = max(length_of_longest_substring, i - starting_index_of_current_substring + 1)
return length_of_longest_substring
## Main
result = {}
for string in ['abcabcbb', 'bbbbb', 'ppwwkew', 'wcabcdeghi', 'bpfbhmipx', 'tmmzuxt', 'AABGAKGIMN', 'stackoverflow']:
result[string] = lengthOfLongestSubstring(string)
print(result)
Sample run:
{'abcabcbb': 3, 'bbbbb': 1, 'ppwwkew': 3, 'wcabcdeghi': 8, 'bpfbhmipx': 7, 'tmmzuxt': 5, 'AABGAKGIMN': 6, 'stackoverflow': 11}

This post is pretty old, but I think my solution fixes the bug in the original code.
def lengthOfLongestSubstring(s):
checklist = {}
starting_index_of_current_substring = 0
length_of_longest_substring = 0
for i, v in enumerate(s):
if v in checklist:
if checklist[v] >= starting_index_of_current_substring:
starting_index_of_current_substring = checklist[v] + 1
length_of_current_substring = i - starting_index_of_current_substring + 1
length_of_longest_substring = max(length_of_current_substring, length_of_longest_substring)
checklist[v] = i
return length_of_longest_substring

This doesnt really iterate upon your solution, but it's a bit simpler approach, just to give you an idea how it could be also solved.
def longest_substr(s):
longest = 0
for start_index in range(len(s)):
contains = set()
for letter in s[start_index:]:
if letter in contains:
break
contains.add(letter)
longest = max(longest, len(contains))
return longest

0
I would prefer this solution>>
Time and Space Management Optimised:
def lengthOfLongestSubstring(self, s: str) -> int:
curlen = maxlen = 0 # curlen saves the max len of substrings ending with current num
for i, num in enumerate(s):
curlen -= s[i-curlen:i].find(num)
maxlen = max(maxlen, curlen)
return maxlen

Find Longest Substring in the string without repeating characters.
def find_non_repeating_substring(input_str):
output_length = 0
longest_sub_str = ''
len_str = len(input_str)
index = 0
while len_str != 1:
l_str = ''
for i in range(index, len(input_str)):
if input_str[i] not in l_str:
l_str = l_str + input_str[i]
else:
break
sub_str_length = len(l_str)
if sub_str_length > output_length:
output_length = sub_str_length
longest_sub_str = l_str
len_str = len_str -1
index = index + 1
return output_length,longest_sub_str
if __name__ == '__main__':
input_str = raw_input("Please enter the string: ")
sub_str_length, sub_str = find_non_repeating_substring(input_str)
print ('Longest Substing lenght is "{0}" and the sub string is "{1}"'.format(sub_str_length, sub_str))```

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)